Question

The CEO of the Jen and Benny's ice cream company is concerned about the net weight of ice cream in their 50 ounce ice cream tubs. He decides that he wants to be fairly sure that the mean weight of these ice cream tubs (μ) is greater than 52 ounces. A hypothesis test is conducted with the following hypotheses: H0: μ = 52 Ha: μ > 52 The level of significance used in this test is α = 0.05. A random sample of 26 ice cream tubs are collected and weighed. The sample mean weight is calculated to be x = 54.98 and the sample standard deviation is s = 8.43. a)Calculate the test statistic (t) for this test. Give your answer to 4 decimal places.

Answer 1

Test statistics = 1.8025

Explanation:

We are given that the CEO of the Jen and Benny's ice cream company is concerned about the net weight of ice cream in their 50 ounce ice cream tubs. He decides that he wants to be fairly sure that the mean weight of these ice cream tubs (μ) is greater than 52 ounces. For this;

Null Hypothesis,
`H_0` :
`\mu` = 52

Alternate Hypothesis,
`H_1` :
`\mu` > 52

Also, a random sample of 26 ice cream tubs are collected and weighed. The sample mean weight is calculated to be x = 54.98 and the sample standard deviation is s = 8.43.

The test statistics used here will be;

T.S. =
`(Xbar -\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where, X bar = sample mean = 54.98

s = sample standard deviation = 8.43

n = sample size = 26

So, test statistics =
`(54.98 -52)/((8.43)/(√(26) ) )` ~
`t_2_5`

= 1.8025

Now, assuming 5% level of significance t table gives critical value of 1.708. Since our test statistics is more than the critical value so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the mean weight of these ice cream tubs (μ) is greater than 52 ounces.

Answer 2

`t=(54.98-52)/((8.43)/(√(26)))=1.8025`

We need to find the degrees of freedom given by:

`df = n-1= 26-1 = 25`

Since is a right tailed test the p value would be:

`p_v =P(t_(25)>1.8025)=0.0418`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.

Explanation:

Data given and notation

`\bar X=54.98` represent the sample mean

`s=8.43` represent the sample deviation

`n=26` sample size

`\mu_o =52` represent the value that we want to test

`\alpha=0.05` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 52, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 52`

Alternative hypothesis:
`\mu > 52`

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(54.98-52)/((8.43)/(√(26)))=1.8025`

P-value

We need to find the degrees of freedom given by:

`df = n-1= 26-1 = 25`

Since is a right tailed test the p value would be:

`p_v =P(t_(25)>1.8025)=0.0418`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.