Question

A group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are listed below. Use a 0.10 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one​ minute?

80

93

46

76

53

31

69

74

74

54

76

76

103

100

76

What are the null and alternative​ hypotheses?

Determine the test statistic. Round to two decimal places.

Determine the P-value. Round to three decimal places.

State the final conclusion that addresses the original claim.

______ H0. There is ______ evidence to conclude that the original claim that the mean of the population of estimated is 60 seconds _______ correct. It ________ that as a group the students are reasonably good at estimating one minute.

Answer 1

The data provided is:

	Data
	80
	93
	46
	76
	53
	31
	69
	74
	74
	54
	76
	76
	103
	100
	76
Count	15
Average	72.0667
SD	19.5903

The provided sample mean is and the sample standard deviation is , and the sample size is

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 60

Ha: μ ≠ 60

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, will be used.

Rejection Region

Based on the information provided, the significance level is α=0.10, and the critical value for a two-tailed test is tc​=1.761.

The rejection region for this two-tailed test is R={t:∣t∣>1.761}

(2) Test Statistics

The t-statistic is computed as follows:

(3)

The p-value is p=0.0317

(4) The decision about the null hypothesis

Reject H0. There is sufficient evidence to conclude that the original claim that the mean of the population of estimated is 60 seconds is not correct. It does not seem that as a group the students are reasonably good at estimating one minute.

Clarification

Since it is observed that ∣t∣=2.386>tc​=1.761, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0317, and since p=0.0317<0.10, it is concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 60, at the 0.10 significance level.

Graphically

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