Answer Save Question 11 (Mandatory) (8 points) Test the claim 75 at a-o.10. A sa

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Answer Save Question 11 (Mandatory) (8 points) Test the claim 75 at a-o.10. A sample of size 76 yields a sample mean of 99 and a sample standard deviation of 3. Select an Option The test statistic 59.7792 lower critical value-1.293, pvalue 1.000, do not reject H0, conclude at least 75 The test statistic 59.7792 upper critical value +1.293, pvalue 0.000, reject H0, conclude 75, The test statistic 59.779 falls outside critical values ± 1.665, pvalue 0.000, reject Ho, 75 Question 12 (Mandatory) (8 points)

Explanation / Answer

Solution:-

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

Null hypothesis: > 75

Alternative hypothesis: < 75

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

Formulate an analysis plan. For this analysis, the significance level is 0.10. 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 = 0.4015

DF = n - 1 = 76 - 1

D.F = 75

t = (x - ) / SE

t = 59.78

tcritical = 1.293

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.

The observed sample mean produced a t statistic test statistic of 59.78

Thus the P-value in this analysis is 0.0000.

Interpret results. Since the t-value (59.78) is greater than the tcritical value (1.293), we cannot reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that > 75.

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