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Lelechka [254]
3 years ago
5

Location is knownLocation is known to affect the number, of a particular item, sold by an automobile dealer. Two different locat

ions, A and B, are selected on an experimental basis. Location A was observed for 18 days and location B was observed for 13 days. The number of the particular items sold per day was recorded for each location. On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4. Does the data provide sufficient evidence to conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B at the 0.01 level of significance
Mathematics
1 answer:
KIM [24]3 years ago
3 0

Answer:

There is  enough evidence to support the claim that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0

Being μ1: true mean number of sales for Location A and μ2: true mean number of sales for Location B.

The significance level is 0.01.

The sample 1 has a size n1=18 and the sample 2 has a size n2=13.

The sample 1 has a mean of 39 and a standard deviation of 8.

The sample 2 has a mean of 49 and a standard deviation of 4.

The difference between sample means is Md=-10.

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{8^2}{18}+\dfrac{4^2}{13}}\\\\\\s_{M_d}=\sqrt{3.556+1.231}=\sqrt{4.786}=2.188

Then, we can calculate the t-statistic as:

The degrees of freedom for this test are:

This test is a left-tailed test, with 29 degrees of freedom and t=-4.571, so the P-value for this test is calculated as (using a t-table):

P-value=P(t

As the P-value (0.00004) is smaller than the significance level (0.01), the effect is significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

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First, we need to find the z-score and this is expressed according to the formula;

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Given the following parameters

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If the true mean is $273,000, then;

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If the true mean is $277,000

z=\dfrac{277,000-275000}{10,500} \\z = \frac{2000}{10500}\\z= 0.1905

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Pr( -0.1095 ≤ z ≤ 0.1095)

According to the z table, the probability that the average selling price of the 35 homes will be within $2000 of the true mean is 0.5753

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Find out the statement true regarding the diagram given in the question

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