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Wize AP Statistics Textbook > Hypothesis Testing for Mean and Proportion

Hypothesis Test for a Mean

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Hypothesis Test for a Mean

A hypothesis test may be performed for a population mean μ\mu. Be sure to review the hypothesis test steps before studying this section.

The test statistic is depends on 3 things:
  1. Whether the population standard deviation σ\sigma is known or not
  2. Whether the population distribution is normal or not
  3. Sample size

→ z\rightarrow\ z (if σ\sigma is known)
→ t\rightarrow\ t (if σ\sigma is unknown)







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Example: Hypothesis Test (Mean) - One-Sided Test


In 2013, the business department reported that the average monthly salary for co-op students is $2,960. Kenya randomly sampled 25 co-op students in the business department. The sample mean is $3,325 with a standard deviation of $940. Assume the population is normally distributed. At the significance level α=0.05\alpha=0.05, is there evidence that the monthly salary is higher now?

(a) Is this a z-test or a t-test?

This is a t-test because σ\sigma is unknown. Although the sample size is not very large, the population is assumed to be normal so we can proceed with the t-test.


(b) State the hypotheses:


Ho: μ=2960H_o:\ \mu=2960
Ha: μ>2960H_a:\ \mu>2960

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(c) What is the critical value?


df=n−1=25−1=24df=n-1=25-1=24
one-tail test ("greater than")
𝜶 = 𝟎.𝟎𝟓


CV=1.711CV=1.711 (t-table)



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(d) Find the test statistic:

t=x‾−μsn=3325−296094025=1.942t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}=\frac{3325-2960}{\frac{940}{\sqrt{25}}}=1.942



(e) Find the p-value.

t-table: 0.025<0.025< p-value <0.05<0.05
Software: p-value = 0.032


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(f) Select the correct conclusion:

i. We reject 𝐻o; there is evidence that the monthly salary is higher now.
ii. We reject 𝐻o; there is no evidence that the monthly salary is higher now.
iii. We fail to reject 𝐻o; there is evidence that the monthly salary is higher now.
iv. We fail to reject 𝐻o; there is no evidence that the monthly salary is higher now.

i. The p-value is smaller than the significance level (0.05). We reject HoH_o; conclude as per HaH_a: there is evidence that the monthly salary is higher now.

You can also say: The test statistic (1.942) is greater than the critical value (1.711). Reject Ho.
There is evidence that the average co-op salary is greater than $2,960.

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(g) What would be the conclusion if α=0.01\alpha=0.01?

i. We reject 𝐻o; there is no evidence that the monthly salary is higher now.
ii. We fail to reject 𝐻o; there is evidence that the monthly salary is higher now.
iii. We fail to reject 𝐻a; there is no evidence that the monthly salary is the same.
iv. We fail to reject 𝐻o; there is no evidence that the monthly salary is higher now.

iv. The p-value is greater than the significance level (0.01). Fail to reject HoH_o. (You do not reject HaH_a!)
There is no evidence that the average co-op salary is greater than $2,960.

If you are curious, the critical value at a = 0.01 is 2.492
The test statistic (1.9415) is smaller than the critical value (2.492). Fail to reject Ho.



(h) What does that say about the strength of the evidence?

Since we can reject Ho at a = 0.05 but not at a = 0.01, we can say that there is strong evidence but not very strong.

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Example: Hypothesis Test (Mean) - Two-Sided Test


The average time it takes to find a parking spot at the mall is 11.5 minutes. However, the mall's property was recently rezoned. Has the average time to find a parking spot changed since the rezoning? Based on a random sample of 41 cars, the average time for shoppers to find a parking spot is 10.25 minutes with a standard deviation of 4.5 minutes. The significance level is α=0.05\alpha=0.05.


(a) Is this a z-test or a t-test?

This is a t-test because σ\sigma is unknown. Although the sample size is not very large, the population is assumed to be normal so we can proceed with the test.


(b) State the hypotheses:

Ho: μ=11.5H_o:\ \mu=11.5
Ha: μ≠ 11.5H_a:\ \mu\neq\ 11.5

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(c) What is the critical value?


df=n−1=41−1=40df=n-1=41-1=40
two-tail test ("changed")
𝜶 = 𝟎.𝟎𝟓


CV=2.021CV=2.021 (t-table)


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(d) Find the test statistic:

t=x‾−μsn=10.25−11.54.541=−1.779t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}=\frac{10.25-11.5}{\frac{4.5}{\sqrt{41}}}=-1.779


(e) Find the p-value. (Remember: this is a two-tail test!)

t-table: 0.05<0.05< p-value <0.10<0.10 (be sure to find the correct p-value based on a two-tail test!)
Software: p-value = 0.0829


(f) Select the correct conclusion:

i. There is evidence that the average time to find a parking spot has decreased.
ii. There is evidence that the average time to find a parking spot has changed.
iii. There is evidence that the average time to find a parking spot has not changed.
iv. There is no evidence that the average time to find a parking spot has changed.


iv. The p-value is greater than the significance level (0.05). We fail to reject HoH_o, then we have no evidence for HaH_a.

(iii. is not correct - we didn't prove that HoH_o is true.)

The marketing department states that the average spending at the store is $335, but they believe that customers are spending less money, on average. A random sample of 31 customers reveal a mean of $302 with a standard deviation of $128.

(i) What are the correct hypotheses?