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Your friend Fred claims that his average gym workouttime is at least 150 minutes per day. You pick a randomsample of 10 days and observe that on the days in thesample, Fred’s average workout time was 136 minutes.The standard deviation of workout times in the samplewas 28 minutes. You plan to use the sample results to testFred’s claim. Assume that Fred’s workout times are normally distributed.

Required:
Set up an appropriate hypothesis test to test Fred's claim. Use a significance level of 5%. Should Fred's claim be rejected? Explain.

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Answer:

Step-by-step explanation:

Given that:

population mean = 150

Sample size  = 10

Sample mean = 136

Standard deviation = 28

Significance level = 0.05

The null hypothesis:

[tex]H_o : \mu \ge 150[/tex]

The alternative hypothesis

[tex]H_a : \mu < 150[/tex]

This is a left-tailed test.

The t-test statistics can be computed as:

[tex]t = \dfrac{\overline x - \mu}{ \dfrac {s} {\sqrt{n}} }[/tex]

[tex]t = \dfrac{136 - 150}{ \dfrac {28} {\sqrt{150}} }[/tex]

t = -1.58

The P-value = P(t < -1.58)

Using the t tables

P-value = 0.0743

Decision rule: To reject the null hypothesis if the P-value is lesser than the level of significance.

Conclusion: We fail to reject the null hypothesis; since P-value is greater than (∝). Thus, there is sufficient evidence to support Fred's Claim.

No, we can not reject Fred's claim.

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