Step 1 of 5: Enter the hypotheses: Step 2 of 5: Enter the value of the z test st

Question

Step 1 of 5: Enter the hypotheses:

Step 2 of 5:

Enter the value of the z test statistic. Round your answer to two decimal places.

Step 3 of 5: Specify if the test is one-tailed or two-tailed.

Step 4 of 5: Enter the decision rule.

Step 5 of 5: Enter the conclusion.

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 437.0 gram setting. Based on a 35 bag sample where the mean is 431.0 grams, is there sufficient evidence at the 0.02 level that the bags are underfilled? Assume the standard deviation is known to be 26.O

Explanation / Answer

Solution:-

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

Null hypothesis: > 437

Alternative hypothesis: < 437

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 0.02. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE) and the z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 4.395

z = (x - ) / SE

z = - 1.37

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 - 1.37. We use the z Distribution Calculator to find P(z < - 1.37) = 0.0853.

Interpret results. Since the P-value (0.0853) is greater than the significance level (0.02), we cannot reject the null hypothesis.

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