A random sample of 77 eighth grade students\' scores on a national mathematics a

Question

A random sample of 77 eighth grade students' scores on a national mathematics assessment test has a mean score of 284 with a standard deviation of 40. This test result prompts a state school administrator to declare that the mean score for the state's eight grades of this exam is more than 280. At alpha = 0.13, is there enough evidence to support the administrator's claim? Complete parts (a) through (e). (a) Write the claim mathematically and identify H_0 and H_a. Choose the correct answer below. A. H_0: mu = 280 (claim) H_a: mu > 280 B. H_0: mu 280 D. H_0: mu = 280 H_a: mu > 280 (claim) E. H_0: mu lessthanorequalto 280 H_a: mu > 280 (claim) F. H_0: mu greaterthanorequalto 280 (claim) H_a: mu

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: < 280
Alternative hypothesis: > 280

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 0.13. 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 z statistic (z).

SE = s / sqrt(n) = 40 / sqrt(77) = 4.55842305839
DF = n - 1 = 77 - 1 = 78
z = (x - ) / SE = (284 - 280)/4.55842305839 = 0.87749643873 or 0.88

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.

Here is the logic of the analysis: Given the alternative hypothesis ( > 280), we want to know whether the observed sample mean is large enough to cause us to reject the null hypothesis.

The observed sample mean produced a t statistic test statistic of 0.88. We use the t Distribution Calculator to find P(t > 0.88)

The P-Value is 0.190135 or 0.190
The result is not significant at p < 0.13.

Interpret results. Since the P-value (0.190) is greater than the significance level (0.13), we cannot reject the null hypothesis.

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