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\[H_{o}: \mu = 32 \\ H_{a}: \mu ≠32 \] b. What is the conclusion of the test when the null is not rejected?
There is not enough statistical evidence to suggest that the average fill weight is different from 32 ounces.
There is enough information to suggest that the true average fill weight is different from 32 ounces.
\[ H_{o}: \mu = 200 \\ H_{a}: \mu < 200 \] a. What is the conclusion of the test when null is not rejected?
There is not enough statistical information to suggest that average costs were reduced from the new manufacturing method.
There is enough statistical informatino to suggest that average costs were reduced as a result of the new manufacturing method.
\[H_{o}:\mu = 25,000 \\ H_{a}: \mu >25,000 \] b. Conceptually, what is the Type I error for this particular problem? What are the consequences of making this error?
The Type I error is a false positive. In this case it would refer to the case where we suggest that the new incentive program increases average sales when it really doesn’t. The consequence would be that the electronics store would likely adopt the new incentive program when it would likely have no effect on increasing sales.
The Type II error is a false negative. In this case it would represent the case where we suggest the new incentive program doesn’t increase sales when in reality the program does. The consequence of making this error is that the electronics store would likely not adopt the new program when it would likely increase sales.
The Type I error for this particular problem would be suggesting that the new method reduces costs when in reality they don’t. The consequence is that the facility would likely adopt the new method to reduce costs while the new method actually won’t reduce costs at all.
The Type II error for this particular problem is to fail to suggest the new method reduces costs when in reality it does. The facility would likely not adopt the new method when it would likely reduce costs.
\[H_{o}: \mu = 1100 \\ H_{a} : \mu < 1100 \]
The test statistic is -1.3 and the p-value associated with a lower-tail test is 0.1.
Since the p-value of 0.097 is greater than 0.05 - the specified level of \(\alpha\) - we fail to reject the null in favor of the alternative. We do not have enough statistical information to suggest that last minute filers receive a lower average tax refund.
\[ CI = 75.38, 102.62\]
The width of the confidence interval will increase if the sample size decreases since there is less information available in the sample.
A study wants to determine if cell phone bills are higher than $100. Specify the null and alternative hypothesis for this particular test. \[H_{o}: \mu=100 \\ H_{a} : \mu >100 \]
Report the test statistic and p-value from this test and interpret these values. Let α = .01. What is the result from the test? Explain.
The test statistic is -1.67 and the p-value is 0.95. The p-value is much grater than the specified level of alpha. We do not reject the null in favor of the alternative. There is not enough information to suggest that the average cell phone bill is more than $100 per month.
\[Ho: \mu = 1200 \\ Ha: \mu >1200 \]
Report the test statistic and p-value. Interpret these values. The test statistic is 2.04 and the p-value for an upper tail test is 0.02. The p-value is approximately 0.
Let α = .05. What is the result of the test? Explain this result.
The p-value is approximately 0. So, we reject the null in favor of the alternative. This suggests that average rental rates in Chicago do exceed $1200 per month.
Specify the appropriate null and alternative hypotheses for this question. \[H_{o}: \mu=9500 \\ H_{a}: \mu >9500 \]
Report the test statistic and p-value. Interpret these values.
The test statistic is 1.37 and the p-value for an upper tail test is 0.09.
Since the p-value is greater than the specified level of \(\alpha\) then we fail to reject the null in favor of the alternative. This suggests that there is no indication that average used car prices are more expensive in the midwest.