Please help with D E FPlease Help with D E F Before the furniture store began its ad campaign, it averaged 130 customers per day. The manager isinvestigating if the average is smaller since the ad came out. The data for the 16 randomly selected dayssince the ad campaign began is shown below:109, 117, 133, 117, 134, 129, 110, 136, 110, 128, 113, 138, 125, 99, 145, 114Assuming that the distribution is normal, what can be concluded at the a = 0.10 level of significance?a. For this study, we should use Select an answerb. The null and alternative hypotheses would be:Но:? vSelect an answer vH1:Select an answer vc. The test statistic(please show your answer to 3 decimal places.)%3Dd. The p-value =(Please show your answer to 4 decimal places.)e. The p-value is ? v af. Based on this, we should Select an answer v the null hypothesis.g. Thus, the final conclusion is that ...O The data suggest that the population mean number of customers since the ad campaign beganis not significantly less than 130 at a = 0.10, so there is insufficient evidence to conclude thatthe population mean number of customers since the ad campaign began is less than 130.O The data suggest the populaton mean is significantly less than 130 at a = 0.10, so there issufficient evidence to conclude that the population mean number of customers since the adcampaign began is less than 130.O The data suggest the population mean is not significantly less than 130 at a = 0.10, so there issufficient evidence to conclude that the population mean number of customers since the adcampaign began is equal to 130.

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Please help with D E F

Please Help with D E F

Before the furniture store began its ad campaign, it averaged 130 customers per day. The manager is
investigating if the average is smaller since the ad came out. The data for the 16 randomly selected days
since the ad campaign began is shown below:
109, 117, 133, 117, 134, 129, 110, 136, 110, 128, 113, 138, 125, 99, 145, 114
Assuming that the distribution is normal, what can be concluded at the a = 0.10 level of significance?
a. For this study, we should use Select an answer
b. The null and alternative hypotheses would be:
Но:
? v
Select an answer v
H1:
Select an answer v
c. The test statistic
(please show your answer to 3 decimal places.)
%3D
d. The p-value =
(Please show your answer to 4 decimal places.)
e. The p-value is ? v a
f. Based on this, we should Select an answer v the null hypothesis.
g. Thus, the final conclusion is that ...
O The data suggest that the population mean number of customers since the ad campaign began
is not significantly less than 130 at a = 0.10, so there is insufficient evidence to conclude that
the population mean number of customers since the ad campaign began is less than 130.
O The data suggest the populaton mean is significantly less than 130 at a = 0.10, so there is
sufficient evidence to conclude that the population mean number of customers since the ad
campaign began is less than 130.
O The data suggest the population mean is not significantly less than 130 at a = 0.10, so there is
sufficient evidence to conclude that the population mean number of customers since the ad
campaign began is equal to 130.

Expert Solution

Step 1

The descriptive statistics is conducted using EXCEL. The software procedure is given below:

Enter the data.
Select Data > Data Analysis >Descriptive Statistics> OK.
Enter Input Range as Data column.
Mark Labels in First Row.
Mark Summary Statistics.
Click OK.

The output using EXCEL is as follows:

From the output, the mean is 122.3125, and the standard deviation is 12.9600.

The test statistic is,

$\begin{array}{rcl} t & = & \frac{\bar{x} - μ}{(\frac{s}{\sqrt{n}})} \\ = & \frac{122.3125 - 130}{(\frac{12.9600}{\sqrt{16}})} \\ = & \frac{- 7.6875}{3.24} \\ = & - 2.373 \end{array}$

The test statistic is -2.373.

The degrees of freedom is,

$\begin{array}{rcl} d f & = & n - 1 \\ = & 16 - 1 \\ = & 15 \end{array}$

The degrees of freedom is 15.

The hypothesis is left tail. The probability of t less than –2.373 with 15 degrees of freedom can be obtained using the excel formula “=T.DIST(–2.373,15,TRUE)”.The probability value is 0.0157.

Thus, the p-value is 0.0157.

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