Mathematics, 24.03.2020 16:42 harlem3166

Trader Moe’s is considering opening a Fayetteville location but their development office has learned that a main competitor, The Whole Market, currently has a store in development here. Trader Moe’s conducts a survey to see which store shoppers would choose if both were available within a 5 mile radius. Out of 200 shoppers surveyed, 73 would go to Trader Moe’s. Assunming the 95% confidence interval for those who would choose Trader Moe’s. What would happen to the confidence interval if a 99% confidence level were used?

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Answer from: kseniyayakimov

95% confidence

$0.365 - 1.96\sqrt{\frac{0.365(1-0.365)}{200}}=0.298$

$0.365 + 1.96\sqrt{\frac{0.365(1-0.365)}{200}}=0.432$

The 95% confidence interval would be given by (0.298;0.432)

99% confidence

$0.365 - 2.58\sqrt{\frac{0.365(1-0.365)}{200}}=0.277$

$0.365 + 2.58\sqrt{\frac{0.365(1-0.365)}{200}}=0.453$

The 99% confidence interval would be given by (0.27;0.453)

So we can conclude that the 99% confidence interval is wider compared to the 95% interval

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The estimated proportion for this cae is given by:

$\hat p =\frac{73}{200}=0.365$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.365 - 1.96\sqrt{\frac{0.365(1-0.365)}{200}}=0.298$

$0.365 + 1.96\sqrt{\frac{0.365(1-0.365)}{200}}=0.432$

The 95% confidence interval would be given by (0.298;0.432)

For the 99% confidence interval, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

If we replace the values obtained we got:

$0.365 - 2.58\sqrt{\frac{0.365(1-0.365)}{200}}=0.277$

$0.365 + 2.58\sqrt{\frac{0.365(1-0.365)}{200}}=0.453$

The 99% confidence interval would be given by (0.27;0.453)

So we can conclude that the 99% confidence interval is wider compared to the 95% interval

Answer from: Quest

Iwould like to think so. what’s the question you need with?

Answer from: Quest

$a^b=\underbrace{a\cdot a\cdot \ldots \cdot a}_{b}$

a - base

b - exponent

$8^3=\underbrace{8\cdot 8 \cdot 8}_{3}$

so,

the base is 8the exponent is 3the exponent tells you to multiply 8 together three timesthe expanded form is 8*8*8

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