A Carpet Manufacturer Is Studying Differences Between Two Of

A Carpet Manufacturer Is Studying Differences Between Two Of Its Major

A carpet manufacturer is studying differences between two of its major outlet stores. The company has decided to revise and improve its operations and response time to customer needs and is particularly interested in the time it takes customers to receive carpeting that was ordered from the plant. A sample of 41 delivery times for the most popular type of carpet for store A revealed an average delivery time of 34.3 days with a standard deviation of 2.4 days. A similar sample of 31 delivery times for store B revealed an average of 43.7 days with a standard deviation of 3.1 days. At the .05 level of significance, is there evidence of a difference in variability in the shipping time between the two outlets? What decision should the carpet manufacturer make? 1 Miranda Meadow, a Virginia senatorial candidate wants an estimate of the proportion of the population who will support her in the November election. Assume a 95% level of confidence. Meadow wants the estimate to be within 0.04 of the true proportion. Meadow’s political advisors have no estimate available for the proportion that will support her. a) How large a sample is required? b) What sampling technique should be used to select such a sample? Explain. c) Past surveys reveal that 10% of the voters in Meadow’s state are willing to spend more than $1,000 on a fundraiser dinner. Meadow wants to update this percentage. The new study is to use the 90 percent confidence level. The estimate is to be within 1 percent of the population proportion. What is the necessary sample size? d) Meadow’s advisors also want to determine the average cost of a 30-second TV ad within the state. What sample size must you take to be 96% confident that the results will be within $75 of the true mean cost per ad given that the mean is $771?

Paper For Above instruction

The study of variability between two outlet stores' delivery times is essential for a carpet manufacturer aiming to enhance operational efficiency. The company collected sample data from two outlets, Store A and Store B. Store A’s sample included 41 delivery times with a mean of 34.3 days and a standard deviation of 2.4 days, while Store B’s sample included 31 delivery times, with a mean of 43.7 days and a standard deviation of 3.1 days. The core question pertains to whether the variability in delivery times differs significantly between the two stores at a 5% significance level.

To determine this, an F-test for equality of variances is appropriate. The null hypothesis (H0) states that the variances are equal, while the alternative hypothesis (Ha) posits that the variances differ. The test statistic is calculated as the ratio of the larger variance to the smaller variance. The variances are obtained by squaring the standard deviations: for Store A, variance is 2.4² = 5.76; for Store B, variance is 3.1² = 9.61. The ratio of variances is 9.61/5.76 ≈ 1.67, with the numerator associated with Store B, which has the larger variance.

Using the F-distribution with numerator degrees of freedom (df1) = 30 (n2 - 1) and denominator degrees of freedom (df2) = 40 (n1 - 1), the critical value at α = 0.05 can be looked up in an F-table or calculated via software. The critical F-value approximates to 1.96 (for a two-tailed test). Since the calculated F-value (1.67) is less than the critical value, there is insufficient evidence to reject the null hypothesis. Therefore, based on this test, the variability in delivery times does not significantly differ between the two outlets at the 5% significance level. The manufacturer might decide that, from a variability standpoint, the two stores are comparable, although the mean differences in delivery times are notable and could also warrant further analysis.

In the context of the second part, Miranda Meadow seeks to estimate the support proportion in her constituency with a specified confidence level and margin of error. When no prior estimate of the support proportion is available, the most conservative approach is to assume p = 0.5, which maximizes the required sample size. Using the formula for sample size for proportions: n = (Z² p (1 - p)) / E², where Z is the Z-score corresponding to the confidence level, p is the estimated proportion, and E is the margin of error. For a 95% confidence level, Z ≈ 1.96; E = 0.04; p = 0.5. Plugging in these values yields n = (1.96² 0.5 0.5) / 0.04² ≈ 600.25, rounded up to 601 respondents. Therefore, a sample of at least 601 voters is necessary.

The sampling technique should be probability sampling, preferably simple random sampling, to ensure each voter has an equal chance of selection, thereby providing an unbiased estimate of the population support proportion.

Next, Meadow's team intends to update the proportion of voters willing to spend more than $1,000 on a fundraiser dinner. They seek a 90% confidence level and a margin of error of 1%. With past data indicating a 10% support rate, p = 0.10. The critical Z-value at 90% confidence is approximately 1.645. Applying the same formula: n = (Z² p (1 - p)) / E², we get n = (1.645² 0.10 0.90) / 0.01² ≈ 1383. Constructing a sufficiently large sample ensures the estimate's precision within the desired margin.

Finally, to determine the average cost of a 30-second TV ad with 96% confidence and a margin of error of $75 (given the known standard deviation of $771), we employ the sample size formula for means: n = (Z σ / E)². The Z-value at 96% confidence is approximately 2.05. Calculating yields n = (2.05 771 / 75)² ≈ (2.05 * 10.28)² ≈ (21.1)² ≈ 445.21, rounded to 446. Therefore, at least 446 samples of ad costs are needed to estimate the mean within the specified margin of error.

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