Question

1. The test for the mean with detailed work shown and detailed discussion. Use the 6-step procedure for hypothesis testing. Explain why you are testing for the numbers in Ho(null hypothesis) for both the populations. A) Test if the mean price for clothing in the entire American Eagle store population, using the American Eagle data, is greater than 69. Conduct the test at the 5% level of significance. B) Do the same for the Forever 21 data with the same levels.

2. Then test for the difference of two means with detailed work shown and detailed discussion of the two data sets. Test if there is a difference between the two population means. Where population one is the clothing from American Eagle (using the data from American Eagle) and population two is the clothing from Forever 21 (using the data from Forever 21).

Using your results in phases one and two now test each population parameter equal to some value. Test if the two parameters are equal. Again provide details of your calculations.

Based on this information:

Price Difference Between Two Popular Clothing Stores																Sum
AmericanEagle	$69	$25	$30	$20	$15	$15	$40	$50	$40	$40	$60	$60	$60	$70	$30	$664
Forever21	$36	$14	$22	$14	$10	$8	$34	$32	$29	$29	$32	$39	$54	$78	$32	$463

Statistics

Variable	Total Count	Mean	StDev	Variance	Median	Mode
American Eagle	15	41.60	19.04	362.69	40.00	40, 60
Forever21	15	30.87	17.88	319.70	32.00	32

Thus the 95% CI is:

Descriptive Statistics

Sample	N	Mean	StDev	SE Mean
AE	15	41.6	19.0	4.9
Forever21	15	30.9	17.9	4.6

Method   

μ₁: population mean of AE

µ₂: population mean of Forever21

Difference: μ₁ - µ₂

            

Equal variances are not assumed for this analysis.      

Estimation For Difference

Difference	95% CI for Difference
10.73	(-3.11, 24.57)

Test                                                               

Null hypothesis			H₀: μ₁ - µ₂ = 0
Alternative hypothesis			H₁: μ₁ - µ₂ ≠ 0
T-Value	DF	P-Value
1.59	27	0.123

Answer 1

Solution X: American eagle clata , u=15 X2: Forever 21 data n=15 Here su = 41.6, si = 19, Standard error of mean Sică,) = 4.9 x2 = 30.9, 82 = 17.9 S₂(T) = 4.6 solution of question O Ho'l=69 USH: M769 so t-stat = zł - MnoN tut, da S,COW t-stat 41.6-69 - 27.U Gig -5.5918 4.9 12 value = PET-5.5910] at Cht) alf=15+) = que = 14dt gggg so we fail to Reject to at £2105 & conclude that we donot have sufficient endence Against Hoi

for - Forever 21 data Holly 69 vs H: M₂ 769 t-stat łz- una ntny, a S₂ (27) t stat 30.9-69 4.6 8.2026 12 value = P{TY-81 2826 at (hy) = (154)= 1 udf 9 9999 so we fail to Reject Hoi because ip-value is greater than a = 105. a sve conclude that mean value of Forever alls hat greater than 69. .Quus 2 ms Hote=12 or ho. Mi-42=0 us H, MU2 con: Mi-M2 70 ar ar ~ tuithaa , & Je2Cth the 2 Here 82= H) 8,²+ M21) 822 lup 19²+(17.9}}} nituz-2 15+15-2 = 340.705 82 1017 so t= 4106-300g 1.5875 6,7399 340.705X( (tstts) so t = 1.59 p-value = P{ (7171.59 at 1, +42-2 af 1123 - 15415-2=28d. 2

Here p-value is greater than d = 105 so we fail to Reject no: UF42 & conclude that mean of AE is not different with a hean of Forever 21. that is a difference is not significantly different nean & we also see that 95% confidence Interval for Mi-M2 L-3111, 24,57 kahich also suggest that there is no difference between the two population mean because confidence Interual contain the value zese.