A random sample of 100 observations from a population with standard deviation 12

Question

A random sample of 100 observations from a population with standard deviation 12.13 yielded a sample mean of 92.3. 1. Given that the null hypothesis is mu = 90 and the alternative hypothesis is mu > 90 using alpha = .05, find the following: (a) Test statistic = (b) P - value: (c) The conclusion for this test is: A. Reject the null hypothesis B. There is insufficient evidence to reject the null hypothesis C. None of the above 2. Given that the null hypothesis is mu = 90 and the alternative hypothesis is mu notequalto 90 using alpha = .05, find the following: (a) Test statistic = (b) P - value: (c) The conclusion for this test is: A. Reject the null hypothesis B. There is insufficient evidence to reject the null hypothesis C. None of the above

Explanation / Answer

Solution:-

1)

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

Null hypothesis: < 90

Alternative hypothesis: > 90

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 0.05. 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 = 1.213

DF = n - 1 = 100 - 1

D.F = 99

t = (x - ) / SE

t = 1.896

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 1.896. We use the t Distribution Calculator to find P(t > 1.896) = 0.03044

Interpret results. Since the P-value (0.03044) is less than the significance level (0.05), we have to reject the null hypothesis.

2)

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

Null hypothesis: = 90

Alternative hypothesis: 90

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 0.05. 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 = 1.213

DF = n - 1 = 100 - 1

D.F = 99

t = (x - ) / SE

t = 1.896

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 99 degrees of freedom is less than -1.896 or greater than 1.896.

Thus, the P-value = 0.06088

Interpret results. Since the P-value (0.06088) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Do not reject the null hypothesis.

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