Question

In: Statistics and Probability

You wish to test the following claim ( H a H a ) at a significance...

You wish to test the following claim ( H a H a ) at a significance level of ? = 0.01 ? = 0.01 . H o : ? = 85.4 H o : ? = 85.4 H a : ? < 85.4 H a : ? < 85.4 You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data: Column A Column B Column C Column D Column E 55.3 67.7 77.8 68.7 77.8 55.3 92.7 99.5 77.8 53.8 60.7 69.2 75.1 101.5 101.5 85.8 67.2 74 92.7 88.7 102.7 74.8 77.4 59.8 75.1 69.7 86.6 75.9 107.5 64.3 87.4 67.7 67.7 91.1 85 92.1 86.6 67.2 86.2 89.6 102.7 73.2 67.2 86.6 65.5 What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) ? ? greater than ? ? This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the population mean is less than 85.4. There is not sufficient evidence to warrant rejection of the claim that the population mean is less than 85.4. The sample data support the claim that the population mean is less than 85.4. There is not sufficient sample evidence to support the claim that the population mean is less than 85.4.

Solutions

Expert Solution

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u > 85.4
Alternative hypothesis: u < 85.4

Note that these hypotheses constitute a one-tailed test.  

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 2.09845
DF = n - 1

D.F = 44
t = (x - u) / SE

t = 3.08

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 3.08.

Thus the P-value in this analysis is 0.004

Interpret results. Since the P-value (0.004) is less than the significance level (0.01), we can reject the null hypothesis.

There is sufficient evidence to warrant rejection of the claim that the population mean is less than 85.4.


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