please help and show work The accounts of a company show that on average, accoun

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please help and show work

The accounts of a company show that on average, accounts receivable are $90.53. An auditor checks a random sample of 49 of these accounts, finding a sample mean of $85.56 and standard deviation of $40.56. Based on these findings, can you conclude the mean accounts receivable is different from $90.53 at a-0.05 For the hypothesis stated above.. Question 1What is the decision rule? Fill in only one of the following statements If the hypothesis is one tailed: Reject H, it f the hypothesis is twvo tailed Reject H, if 29 30 32 34 Question 2 What is the test statistic? Question 3What is the p value? Fill in only one of the following statements t the Z table is appropniate, p-value

Explanation / Answer

Solution:-

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

Null hypothesis: = 90.53

Alternative hypothesis: 90.53

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 = 5.79

DF = n - 1 = 49 - 1

D.F = 48

t = (x - ) / SE

t = - 0.86

tcritical = 1.677(one tailed)

tcritical = 2.011(Two tailed)

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 49 degrees of freedom is less than - 0.86 or greater than 0.86.

Thus, the P-value = 0.394

Interpret results. Since the P-value (0.394) is greater than the significance level (0.05), we cannot reject the null hypothesis.

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