Question

Question 13 2 pts

(CO6) Supplier claims that they are 95% confident that their products will be in the interval of 20.45 to 21.05. You take samples and find that the 95% confidence interval of what they are sending is 20.50 to 21.10. What conclusion can be made?

The supplier is more accurate than they claimed

The supplier products have a lower mean than claimed

The supplier products have a higher mean than claimed

The supplier is less accurate than they have claimed

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Question 14 2 pts

(CO6) In a sample of 17 small candles, the weight is found to be 3.72 ounces with a standard deviation of 0.963 ounces. What would be the 87% confidence interval for the size of the candles?

(3.369, 4.071)

(3.199, 4.241)

(3.347, 4.093)

(2.757, 4.683)

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Question 15 2 pts

(CO6) In a situation where the sample size was increased from 29 to 39, what would be the impact on the confidence interval?

It would become wider due to using the t distribution

It would become narrower with fewer values

It would remain the same as sample size does not impact confidence intervals

It would become narrower due to using the z distribution

Question 23 2 pts

(CO7) A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 720 hours. A random sample of 51 light bulbs as a mean of 705.4 hours with a standard deviation of 62 hours. At an α=0.05, can you support the company’s claim using the test statistic?

Claim is the null, reject the null and cannot support claim as test statistic (-1.68) is in the rejection region defined by the critical value (-1.645)

Claim is the alternative, fail to reject the null and cannot support claim as the test statistic (-1.68) is in the rejection region defined by the critical value (-1.96)

Claim is the alternative, reject the null and support claim as test statistic (-1.68) is not in the rejection region defined by the critical value (-1.96)

Claim is the null, fail to reject the null and cannot support claim as test statistic (-1.68) is not in the rejection region defined by the critical value (-1.645)

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Question 24 2 pts

(CO7) A restaurant claims the customers receive their food in less than 16 minutes. A random sample of 39 customers finds a mean wait time for food to be 15.8 minutes with a standard deviation of 0.7 minutes. At α = 0.04, can you support the organizations’ claim using the test statistic?

Claim is the alternative, reject the null so support the claim as test statistic (-1.78) is in the rejection region defined by the critical value (-1.75)

Claim is the null, reject the null so cannot support the claim as test statistic (-1.78) is in the rejection region defined by the critical value (-2.05)

Claim is the alternative, fail to reject the null so cannot support the claim as test statistic (-1.78) is not in the rejection region defined by the critical value (-2.05)

Claim is the null, fail to reject the null so support the claim as test statistic (-1.78) is not in the rejection region defined by the critical value (-1.75)

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Question 25 2 pts

(CO7) A manufacturer claims that their calculators are 6.800 inches long. A random sample of 55 of their calculators finds they have a mean of 6.812 inches with a standard deviation of 0.05 inches. At α=0.08, can you support the manufacturer’s claim using the p value?

Claim is the alternative, fail to reject the null and support claim as p-value (0.075) is less than alpha (0.08)

Claim is the null, fail to reject the null and support claim as p-value (0.038) is greater than alpha (0.08)

Claim is the alternative, reject the null and cannot support claim as p-value (0.038) is greater than alpha (0.08)

Claim is the null, reject the null and cannot support claim as p-value (0.075) is less than alpha (0.08)

Answer 1

Solution:-

25)

Claim is the null, reject the null and cannot support claim as p-value (0.075) is less than alpha (0.08).

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

Null hypothesis: u = 6.800
Alternative hypothesis: u 6.800

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 = 0.00674

DF = n - 1

D.F = 54
t = (x - u) / SE

t = 1.78

where s is the standard deviation of the sample, x is the sample mean, u 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 54 degrees of freedom is less than -1.78 or greater than 1.78.

Thus, the P-value = 0.0751.

Interpret results. Since the P-value (0.0751) is less than the significance level (0.08), we have to reject the null hypothesis.