4.6 Hypothesis Testing (Chi-square & t-test)

A factory, producing plastic gifts for a fast food restaurant’s Jolly meals, claims that just 1% of the toys produced are faulty.

A restaurant manager wants to test this claim. A box of 200 toys is delivered to the restaurant. The manager checks all the toys in this box and four toys are found to be faulty.

(a) Identify the type of sampling used by the restaurant manager. 

The restaurant manager performs a one-tailed hypothesis test, at the 10% significance level, to determine whether the factory’s claim is reasonable. It is known that faults in the toys occur independently.

(b) Write down the null and alternative hypotheses. 

(c) Find the `p`-value for the test. 

(d) State the conclusion of the test. Give a reason for your answer. 

(a) Convenience

(b) `H_0:` 1% of the toys produced are faulty

`H_1:` More than 1% are faulty

(c) `X ∼ B(200,0.01)`

`P(X >= 4) = 0.142`

(d) 14% > 10%

So there is insufficient evidence to reject `H_0`.

At Springfield University, the weights, in kg, of 10 chinchilla rabbits and 10 sable rabbits were recorded. The aim was to find out whether chinchilla rabbits are generally heavier than sable rabbits. The results obtained are summarized in the following table.

A `t`-test is to be performed at the 5% significance level.

(a) Write down the null and alternative hypotheses. 

(b) Find the `p`-value for this test. 

(c) Write down the conclusion to the test. Give a reason for your answer. 

(a) (let `mu_c` = population mean for chinchilla rabbits, `mu_s` = population mean for sable rabbits)

`H_0: mu_c=mu_s`

`H_1: mu_c > mu_s`

(b)  `p`-value = 0.0408 (0.0408065...)

(c) 0.0408 `<` 0.05.

(there is sufficient evidence to) reject (or not accept) `H_0`

(there is sufficient evidence to suggest that chinchilla rabbits are (generally) heavier than sable rabbits)

Leo is investigating whether a six-sided die is fair. He rolls the die 60 times and records the observed frequencies in the following table:

Leo carries out a `chi^2` goodness of fit test at a 5% significance level.

(a) Write down the null and alternative hypotheses. 

(b) Write down the degrees of freedom. 

(c) Write down the expected frequency of rolling a 1. 

(d) Find the `p`-value for the test. 

(e) State the conclusion of the test. Give a reason for your answer. 

(a) `H_0`: The die is fair

`H_1`: The die is not fair

(b) 5

(c) 10

(d) (`p`-value =) 0.287 (0.28724163...)

(e) 0.287 `>` 0.05

Insufficient evidence to reject the null hypothesis

A group of 1280 students were asked which electronic device they preferred. The results per age group are given in the following table.

(a) A student from the group is chosen at random. Calculate the probability that the student 

(i) prefers a tablet. 

(ii) is 11–13 years old and prefers a mobile phone. 

(iii) prefers a laptop given that they are 17–18 years old. 

(iv) prefers a tablet or is 14–16 years old. 

A `chi^2` test for independence was performed on the collected data at the 1% significance level. The critical value for the test is 13.277.

(b) State the null and alternative hypotheses. 

(c) Write down the number of degrees of freedom. 

(d) 

(i) Write down the `chi^2` test statistic. 

(ii) Write down the `p`-value. 

(iii) State the conclusion for the test in context. Give a reason for your answer.

(a)

(i) `560/1280 \ (7/16, 0.4375)`

(ii) `72/1280 \ (9/160, 0.05625)`

(iii) `153/348 \ (51/116, 0.439655 ...)`

(iv) `160+224+128+205+131`

`848/1280 \ (53/80, 0.6625)`

It is claimed that a new remedy cures 82% of the patients with a particular medical problem.

This remedy is to be used by 115 patients, and it is assumed that the 82% claim is true.

(a) Find the probability that exactly 90 of these patients will be cured. 

(b) Find the probability that at least 95 of these patients will be cured. 

(c) Find the variance in the possible number of patients that will be cured. 

The probability that at least `n` patients will be cured is less than 30%.

(d) Find the least value of `n`. 

A clinic is interested to see if the mean recovery time of their patients who tried the new remedy is less than that of their patients who continued with an older remedy. The clinic randomly selects some of their patients and records their recovery time in days. The results are shown in the table below.

The data is assumed to follow a normal distribution and the population variance is the same for the two groups. A `t`-test is used to compare the means of the two groups at the 10% significance level.

(e) State the appropriate null and alternative hypotheses for this `t`-test. 

(f) Find the `p`-value for this test. 

(g) State the conclusion for this test. Give a reason for your answer. 

(h) Explain what the `p`-value represents. 

(a)  recognition of binomial distribution

e.g. `X ∼ B (115,0.82)`

(`P`(`X` = 90) =) 0.0535 (0.0535325 ... )

(b)  selecting correct region of distribution

e.g. `P(X >= 95)`

0.491 (0.491036...)

(c)  substitution in the variance formula for binomial distribution

`115 × 0.82 × 0.18`

17.0 (16.974)

(d)  attempt to write an expression containing `n` inside the brackets of `P`()

including 0.3 or 0.7

`P(X >= n) < 0.3`

`n` = 98

(e)  (`mu_1`: population mean recovery time for new remedy)

(`mu_2`: population mean recovery time for old remedy)

`H_0`: `mu_1` = `mu_2` (`H_0`: `mu_1` − `mu_2` = 0)

`H_1`: `mu_1` `<` `mu_2` (`H_1`: `mu_1` − `mu_2` `<` 0)

(f)  0.0620 (0.0620061...)

(g)  0.0620 `<` 0.1

(sufficient evidence to) reject `H_0`

(h)  the probability of obtaining results (at least as extreme) as those observed given that the null hypothesis is true

On 90 journeys to his office, Isaac noted whether or not it rained. He also recorded his journey time to the office, and classified each journey as short, medium or long.

Of the 90 journeys to the office, there were 3 short journeys when it rained, 22 medium journeys when it rained, and 15 long journeys when it rained. There were also 14 short journeys when it did not rain.

Isaac carried out a `chi^2` test at the 5% level of significance on these data, looking at the weather and the types of journeys.

(a) Write down `H_0`, the null hypothesis for this test. 

(b) Find the expected number of short trips when it rained. 

The `p`-value for this test is 0.0206.

(c) State the conclusion to Isaac’s test. Justify your reasoning. 

(a) type of journey and whether it rained are independent

(b) `{17 times 40}/90`

`7.56 \ (7.55555..., 68/9)`

(c) type of journey and whether it rained are not independent

0.0206 `<` 0.05

Casanova restaurant offers a set menu where a customer chooses one of the following meals: pasta, fish or shrimp.

The manager surveyed 150 customers and recorded the customer’s age and chosen meal. The data is shown in the following table.

A `chi^2` test was performed at the 10% significance level. The critical value for this test is 4.605.

(a) State `H_0`, the null hypothesis for this test. 

(b) Write down the number of degrees of freedom. 

(c) Show that the expected number of children who chose shrimp is 31, correct to two significant figures. 

(d) Write down 

1. the `chi^2` statistic; 
2. the `p`-value. 

(e) State the conclusion for this test. Give a reason for your answer. 

(f) A customer is selected at random. 

1. Calculate the probability that the customer is an adult. 
2. Calculate the probability that the customer is an adult or that the customer chose shrimp.
3. Given that the customer is a child, calculate the probability that they chose pasta or fish. 

(a) (`H_0`): choice of meal is independent of age (or equivalent)

(b) 2

(c) `{69 times 67}/150`

30.82 (30.8)

31

(d)

(i) (`chi_{calc}^2` =) 2.66 (2.657537)

(ii) (p-value =) 0.265 (0.264803) 

(e) 2.66 `<` 4.605

the null hypothesis is not rejected

(f)

(i) `81/150 \ (27/50, 0.54,54%)`

(ii) `116/150 \ (58/75, 0.773, 0.773333 ... , 77.3%)`

(iii) `34/69 \ (0.493, 0.492753 ... , 49.3%)`

A survey was conducted on a group of people. The first question asked how many pets they each own. The results are summarized in the following table.

(a) Write down the total number of people, from this group, who are pet owners. 

(b) Write down the modal number of pets. 

(c) For these data, write down 

1. the median number of pets;
2. the lower quartile;
3. the upper quartile.

The second question asked each member of the group to state their age and preferred pet. The data obtained is organized in the following table.

(d) Write down the ratio of teenagers to non-teenagers in its simplest form. 

A `chi^2` test is carried out at the 10% significance level.

(e) State

1. the null hypothesis;
2. the alternative hypothesis.

(f) Write down the number of degrees of freedom for this test. 

(g) Calculate the expected number of teenagers that prefer cats. 

(h) Use your graphic display calculator to find the p-value for this test. 

(i) State the conclusion for this test. Give a reason for your answer. 

Abhinav carries out a `chi^2` test at the 1% significance level to determine whether a person’s gender impacts their chosen professional field: engineering, medicine or law. He surveyed 220 people and the results are shown in the table.

(a) State the null hypothesis, `H_0`, for this test. 

(b) Calculate the expected number of male engineers. 

(c) Find the `p`-value for this test. 

Abhinav rejects `H_0`.

(d) State a reason why Abhinav is incorrect in doing so. 

(a) gender and chosen profession are independent

(b) `110/220 times 90/220 times 220 \ ({110 times 90}/220) =45`

(c) 0.0193 (0.0192644...)

(d) the `p`-value is greater than the significance level (1%) 

(i) The number of hours a student spends studying for a particular examination is `x`. The score out of 100 the student receives is `y`. Pairs of (`x,y`) values are recorded for a class of students and the relationship between `x` and `y` is investigated. The results may be summarized in the following table.

The covariance of `x` and `y` is equal to 36.

(a) Find the equation of the least squares regression line of `y` on `x`, expressing your answer in the form `y=mx+c` 

(b)

1. Use your answer to part (a) to predict how many marks a student who studies for 20 hours would achieve. 
2. A teacher wishes to explain to students why they cannot guarantee a score of 100 by studying for the hours calculated in part (b) (i). In order to do so, the value of the product-moment correlation coefficient, `r`, is to be used. 
   1. Calculate `r` for the given data.
   2. Based on the value of 𝑟 obtained, how reliable is the prediction of part (b) (i)? 

(ii) Residents of a small town have savings which are normally distributed with a mean of $3000 and a standard deviation of $500.

(a) 

1. What percentage of townspeople have savings greater than $3200?
2. Two townspeople are chosen at random. What is the probability that both of them have savings between $2300 and $3300? 
3. The percentage of townspeople with savings less than `d` dollars is 74.22%. Find the value of `d`. 

(b) The local newspaper claims that the mean savings is less than $3000. To test this claim, they take a random sample of 100 townspeople and they find that the mean of this sample is $2950. 

1. State the null hypothesis and the alternative hypothesis.
2. Show there is not sufficient evidence to support the newspaper’s claim at the 5% level of significance. 
3. Use the above random sample of 100 townspeople to calculate the 99 % confidence interval for the mean savings. 

(i)

(a) `y - bar y = s_{xy}/s_x^2(x-bar x)`

`y-60=36/3^2(x-10)=>y=4x+20`

(b)

1. `x=20 => y = 4 times 20+20=100`
2. 
   
   1. `r=s_{xy}/{s_x s_y}=36/{3(15)}=0.8`
   2. The value of r indicates fairly good, (or equivalent) not exceptionally (moderately, fairly) strong linear correlation. Thus `x = 20` does not guarantee `y=100` (is not reliable). (Use discretion for equivalent meanings.) 

(ii)

(a)

1. `P`(`X` > 3200) = `P`(`Z` > 0.4) = 1 − 0.6554 = 34.5% (0.345)
2. `P`(2300 < `X` < 3300) = `P`(−1.4 < `Z` < 0.6) = 0.4192 + 0.2257 = 0.645 `=>` `P`(both) = 0.645² = 0.416
3. 0.7422 = `P`(`Z` < 65)
   `{d-3000}/500=0.65 => d` = $3325 (= $3330 to 3s.f.) (accept $3325.07)    

(b)

1. `H_0`: The mean savings is $3000 (`H_0`: `mu` = 3000)
   `H_1`: The mean savings is less than $3000 (`H_1`: `mu` < 3000)
2. `p` = 0.1587
   Since 0.1587 > 0.05, we accept the null hypothesis. (or fail to reject). Therefore, there is not sufficient evidence to support the newspaper’s claim. 
3. 2950 − 2.58(50) < `mu` < 2950 + 2.58(50)
   2.58 `times` 50 = 129
   99% C.I. is 2950 ± 129 (2821, 3079) or (2820, 3080)