Asimple random sample of size n 5 is drawn from a population that is normally di

Question

Asimple random sample of size n 5 is drawn from a population that is normally distributed. The sample mean is found to be x 26.9 and the sample standard deviation is found to be s 6.3. Determine if the population mean is different from 24 at the a 0.01 level of significance. Complete parts (a) through (d) below. (a) Determine the null and alternative hypotheses. Ho: 24 24 (b) Calculate the P-value. P-value (Round to three decimal places as needed.) (c) State the conclusion for the test. O A. Reject Ho because the P-value is greater than the a 0.01 level of significance. O B. Do not reject Ho because the P-value is greater than the 0.01 level of significance. O C. Reject Ho because the P-value is less than the a -0.01 level of significance. O D. Do not reject Ho because the P-value is less than the o 0.01 level of significance

Explanation / Answer

Solution:-

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

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

Null hypothesis: = 24
Alternative hypothesis: 24

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

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) = 6.3 / sqrt(15) = 1.62665
DF = n - 1 = 15 - 1 = 14
t = (x - ) / SE = (26.9 - 24)/1.62665 = 1.7828

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

Since we have a two-tailed test, the P-value is the probability that the t statistic having 14 degrees of freedom is less than -1.7828 or greater than 1.7828.

We use the t Distribution Calculator to find P(t < 1.7828)

The P-Value is 0.096308.
The result is not significant at p < 0.01.

Interpret results. Since the P-value (0.096) is greater than the significance level (0.01), we cannot reject the null hypothesis.

Conclusion. Do not reject Ho. Because P-value is greater than level of significance.

Conclusion in context. There is not sufficient evidence that at 0.01 level of significance, the population mean is different from 24.

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