Question

a grocery store owner wants to know if a new marketing campaign will

Take+home+Exam+1-SPR2019) Mailings ReviewView Normal No Spacing Heading 1 Heading 2 A grocery store owner wants to know if a new marketing campaign will improve the average number of customers who visit the store on a given day. The owner knows that historically they average 490.2 customers per day. After the campaign the owner randomly samples 28 days and finds an average of 512.1 customers with a standard deviation of 83.2. Did the new campaign increase the average number of customers who visit the grocery store? 1. Identify the appropriate test, alpha, and check assumptions Two sample t test Alpha 05 Assumptions means are the same 2. Formulate hypotheses 3. Perform calculations (test statistic & P-value) 4. Compare P-value to alpha and state conclusion 5. Write your conclusion in English

Answer 1

Q2.

i) Since the sample size is less than 30 hence right tail test is applied here

Alpha=0.05= Level of significance

Assumption:

i) Randomly selected Sample

ii) Since it is from a large population it is approximately normal

ii) n<30 then it follows t distribution

ii) The hypotheses are:

$Ho:\mu=490.2$

$Ha:\mu\neq490.2$

3) Test Statistic

$T=\frac{\bar{X}-\mu}{\frac{S}{\sqrt{n}}}=\frac{512.1-490.2}{\frac{83.2}{\sqrt{28}}}=1.393$

P value :

P value associated with T statistic and Degree of freedom=n-1=28-1=27 as 0.088

4). Since P value =0.088 > 0.05( Level of significance) hence we fail to reject the null hypothesis.

5) Conclusion:

Since we fail to reject the Ho (Null hypothesis), then we conclude that we do not have enough evidence to support the claim that the new campaign has increased the average no of customers.

Q3.

One sample Z tail proportion test is applied here.

2) The hypotheses are:

$Ho: P=0.08$

$Ho: P<0.08$

Hence it follows left tail test for proportions

3) Test statistics

$Z=\frac{\hat{P}-P}{\sqrt{\frac{P(1-P)}{n}}}=\frac{\frac{24}{341}-0.08}{\sqrt{\frac{0.08(1-0.08)}{341}}}=-0.0655$

P value:

P value associated with the Z score as 0.258 can be computed from Z statistic table or by the calculator.

4) The P value=0.258>0.05(Alpha) Hence we fail to reject the null hypothesis.

5) Conclusion:

Since we fail to rejctthe null(Ho) Hypothesis hence we conclude that we do not have enough evidence to support the claim.

Q4.

1) The Level of significance =0.01( Alpha)

Assumption :

i) The sample was taken from a large population.

ii) n>30 hence it follows a normal distribution

iii) The sample is randomly selected

2) The hypotheses are:

$Ho:\mu=2400$

$Ha:\mu>2400$

Hence Right tail T-test is used here since sample standard deviation is given

3) Test statistic

$T=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}}=\frac{2500-2400}{\frac{125}{\sqrt{100}}}=8.0$

P value:

P value associated with the T statistics as almost 0.00001=0

4) Since the P value=0< 0,10 hence we reject the null hypothesis .

5) Since we reject the null hypothesis we conclude that we do have enough evidence to support that their mean was higher than of competitors.