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If $n = 15, x ̄ (x - b a r) = 38, a n d s = 10$ , find the margin of error at a 95% confidence level

Answer : Given : Sample mean $(x \bar{)} = 38$ Sample standard deviation $(s) = 10$ Sample size $(n) = 15$ Confidence level $= 95$ Solution : The significance level : $α = 1 - 0.95 = 0.05$ The degrees of…

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Construct a 95% confidence interval for the population mean if the standard deviation of the population is 12, the sample size is 70 and the sample mean is 100.

Answer : Given : Sample mean $(x \bar{)} = 100$ Population standard deviation $(σ) = 12$ Sample size $(n) = 70$ Confidence level $= 95$ Solution : The level of significance : $α = 1 - 0.95 = 0.05$ Critical value…

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Construct a 95% confidence interval to estimate the population mean using the data below: $x ̄ = 48, σ = 10, n = 50$ With 95% confidence, when $n = 50$ , the population mean is between a lower limit of _ and an upper limit of _.

Answer : Given : Sample mean $(x \bar{)} = 48$ Population standard deviation $(σ) = 10$ Sample size $(n) = 50$ Confidence interval level $(C I) = 95$ Solution : The level of significance is calculated as: $$\alpha = 1 – 0.95…

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A professional employee in a large corporation receives an average of $μ = 41.7$ e-mails per day. An anti-spam protection program was installed in the company’s server, and one month later, a random sample of 45 employees showed that they were receiving an average of $x ̄ = 36.2$ e-mails per day. Assume that $σ = 18.45$ . Use a 5% level of significance to test whether there has been a change (either way) in the average number of emails received per day per employee.

a.) What is α? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? $α$ = ___ $H ₀ : μ$ = ___ $H ₁ : μ$ ___The test is a ___ test. b.) Identify the Sampling Distribution you will use. What…

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A sports league introduced a video referee to review certain critical plays. However, the system has not been without its controversies. One particular unfortunate incident led to an outcry among fans. Nevertheless, a poll after this event found that 57% of fans were still in favor of the continued use of the video referee. Suppose the size of this sample was 300 people. Complete parts a and b.

(a) Estimate the true proportion of fans in favor of keeping the video referee using a 99% confidence interval. (b) What is the margin of error for this sample? Answer : Given : Sample proportion $(p \hat{)} = 0.57$ Sample size…

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A random sample of 25 life insurance policyholders showed that the average premium they pay on their life insurance policies is 534 per year with a standard deviation of 63.

Assuming that the life insurance policy premiums for all life insurance policyholders have a normal distribution, make a 99% confidence interval for the population mean, μ. Answer : Given : Sample mean $(x \bar{)} = 534$ Sample standard deviation $(s) =…

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You wish to test the following claim (Ha) at a significance level of α = 0.10.

$H_{o} : μ = 67.9$ $H_{a} : μ \neq 67.9$ You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n = 13 with mean M = 73 and a standard…

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Consider the hypothesis test below. $H_{0} : p_{1} - p_{2} \leq 0$ $H_{a} : p_{1} - p_{2} > 0$

The following results are for independent samples taken from the two populations. $Sample 1 n_{1} = 100 {\overset{―}{p}}_{1} = 0.28$ $Sample 2 n_{2} = 300 {\overset{―}{p}}_{2} = 0.16$ Use pooled estimator of p. a) what is…

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A manufacturer claims that the mean lifetime of its lithium battery is 1000 hours. A homeowner selects 40 batteries and finds the mean lifetime to be 990 hours with a standard deviation of 80 hours. Test the manufacturer’s claim. Use α=0.05.

Answer: Given Data : Hypothesized Population Mean $(μ) = 1000$ Sample Standard Deviation $(s) = 80$ Sample Size $(n) = 40$ Sample Mean $(x \bar{)} = 990$ Significance Level $(α) = 0.05$ The null and alternative hypothesis : $H_{0} : μ = 1000$ $H_{a} : μ! = 1000$ The test statistic : $$t = \frac{\overline{x} – \mu}{\frac{s}{\sqrt{n}}}…

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The amount of weight a high school student carries in his or her bookbag follows a normal distribution with a standard deviation of σ = 2 pounds. Suppose a random sample of 20 bookbags produced a mean of 13 pounds. Construct a 95% confidence interval to estimate the mean bookbag weight for all high school students.

Given Data : Sample mean $(x \bar{)} = 13$ Population standard deviation $(σ) = 2$ Sample size $(n) = 20$ Confidence interval level $(C I) = 95$ The level of significance: $α = 1 - 0.95 = 0.05$ The critical value: $z_{c} = Z_{\frac{α}{2}} = Z_{\frac{0.05}{2}} = 1.96$ The confidence interval: $$CI = \overline{x} \pm z_c \cdot…