Question

In: Statistics and Probability

A group of students estimated the length of one minute without reference to a watch or​...

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 0.05 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? 70 81 42 63 40 27 57 66 63 50 62 73 95 92 62

Solutions

Expert Solution

2 Tailed t test, Single Mean

Given: = 60 seconds

From the Given Data: = 62.87 sec, s = 18.554 seconds, n = 15, = 0.05

The Hypothesis:

H0: = 60: The mean time referenced by students without looking at a watch or clock is equal to 60 seconds.

Ha: 60: The mean referenced by students without looking at a watch or clock is not equal to 60 seconds.

This is a 2 tailed test

The Test Statistic: Since the population standard deviation is unknown, we use the students t test.

The test statistic is given by the equation:

t observed = 0.18

The p Value: The p value (2 tailed) for t = 0.18, for degrees of freedom (df) = n-1 = 14, is; p value = 0.8597

The Critical Value: The critical value (2 Tail) at = 0.05, for df = 14, tcritical= +2.145 and -2.145

The Decision Rule:  

The Critical Value Method: If tobserved is > tcritical or if tobserved is < -tcritical, Then reject H0.

The p-value Method: If P value is < , Then Reject H0.

The Decision:

The Critical Value Method: Since tobserved (0.18) is in between +2.145 and -2.145, We Fail to Reject H0.

The p-value Method: Since P value (0.8597) is > (0.05) , We Fail to Reject H0.

The Conclusion: There is not sufficient evidence at the 95% significance level to warrant rejection of the claim that the mean time referenced by students without looking at a watch or clock is equal to 60 seconds.

_____________________________________________________________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)2.

# X Mean (x - mean)2
1 70 62.87 50.84
2 81 62.87 328.70
3 42 62.87 435.56
4 63 62.87 0.02
5 40 62.87 523.04
6 27 62.87 1286.66
7 57 62.87 34.46
8 66 62.87 9.80
9 63 62.87 0.02
10 50 62.87 165.637
11 62 62.87 0.757
12 73 62.87 102.617
13 95 62.87 1032.337
14 92 62.87 848.557
15 62 62.87 0.757
n 15
Sum 943
Average 62.87
SS(Sum of squares) 4819.7335
Variance = SS/n-1 344.267
Std Dev=Sqrt(Variance) 18.554

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