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Neko [114]
2 years ago
8

Foot locker uses sales per square foot as a meaure of store productivity. sales are currently running at an annual rate of $406

per square foot. you have been asked by management to conduct a study of a sample of 64 foot locker stores.Assume the standard deviate in annual sales per square foot for the population of all 3400 Foot locker stores is $80.
a.) show the sampling distribution of x bar, the sample mean annual sales per square foot for a sample of 64 Foot locker stores.

b.)What is the probability that the sample mean will be within $15 of the population mean?

c.) Suppose you find a sample mean of $380. what is the probability of finding a sample mean of $380 or less? would you consider such a sample to be unusually low performing group of stores?
Mathematics
1 answer:
kotykmax [81]2 years ago
8 0

Answer and explanation:

Given : Foot locker uses sales per square foot as a measure of store productivity. sales are currently running at an annual rate of $406 per square foot. you have been asked by management to conduct a study of a sample of 64 foot locker stores.Assume the standard deviate in annual sales per square foot for the population of all 3400 Foot locker stores is $80.

To find :

a) Show the sampling distribution of x bar, the sample mean annual sales per square foot for a sample of 64 Foot locker stores.

By the central limit theorem also specifies that \bar{x} will have the same expected value as the population mean, that is E(\bar{x}) = 406

The standard deviation of \bar{x} is given by,

\sigma_{\bar{x}}=\frac{\sigma}{\sqrt n}

\sigma_{\bar{x}}=\frac{80}{\sqrt {64}}

\sigma_{\bar{x}}=\frac{80}{8}

\sigma_{\bar{x}}=10

b) What is the probability that the sample mean will be within $15 of the population mean?

The probability is  given by, P(-15

P(\frac{-15}{\sigma_{\bar{x}}}

P(\frac{-15}{10}

P(-1.5

From standard normal table,

P(-1.5

c) Suppose you find a sample mean of $380. what is the probability of finding a sample mean of $380 or less? would you consider such a sample to be unusually low performing group of stores?

P(\bar{x}

 P(\bar{x}

 P(\bar{x}

 P(\bar{x}

From standard normal table,

  P(\bar{x}

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if the sales tax rate is 7.25% in california, then how much would u pay in los angeles for a coat that cost $120.00
natima [27]

Answer:

The amount to be pay for coat is <u>$128.70</u>.

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Given:

The sales tax rate is 7.25% in California.

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Now, to find amount to be paid.

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So, amount to be paid = $120 + 7.25% of $120.00.

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2 years ago
Shanelle already has $137 from babysitting. She needs at least $625 for a new tablet. Which inequality best represents this scen
kap26 [50]

Answer:

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15 members of the Drama Club can make set decorations for their musical in 12 hours. After 4 hours have passed, a few teachers c
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5 teachers

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4 0
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A math teacher tells her students that eating a healthy breakfast on a test day will help their brain function and perform well
kogti [31]

Answer:

Step-by-step explanation:

Hello!

To test the claim that eating a healthy breakfast improves the performance of students on their test a math teacher randomly asked 46 students what did they have for breakfast before they took the final exam and classified them as:

<u>Group 1</u>: Ate healthy breakfast

X₁: Number of students that ate a healthy breakfast before the exam and earned 80% or higher.

n₁= 26

<u>Group 2: </u>Did not eat healthy breakfast

X₂: Number of students that did not eat a healthy breakfast before the exam and earned 80% or higher.

n₂= 20

After the test she counted the number of students that got 80% or more in the test for each group obtaining the following sample proportions:

p'₁= 0.50

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The parameters of study are the population proportions, if the claim is true then p₁ > p₂

And you can determine the hypotheses as

H₀: p₁ ≤ p₂

H₁: p₁ > p₂

α: 0.05

Z= \frac{(p'_1-p'_2)-(p_1-p_2)}{\sqrt{p'(1-p')[\frac{1}{n_1} +\frac{1}{n_2}] } } }≈N(0;1)

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Z_{H_0}= \frac{(0.5-0.4)-0}{\sqrt{0.46(1-0.46)[\frac{1}{26} +\frac{1}{20}] } } }= 0.67

p-value: 0.2514

The decision rule is:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The p-value: 0.2514 is greater than the significance level 0.05, the test is not significant.

At a 5% significance level you can conclude that the population proportion of math students that obtained at least 80% in the test and had a healthy breakfast is equal or less than the population proportion of math students that obtained at least 80% in the test and didn't have a healthy breakfast.

So having a healthy breakfast doesn't seem to improve the grades of students.

I hope this helps!

5 0
2 years ago
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