Question

Help me answer each question STEP by STEP (can use excel, if so show me the formulas please!)

According to a retail organization, holiday shoppers spent an average of $935.58 over the Thanksgiving weekend in 2016. The accompanying data show the amount spent by a random sample of holiday shoppers during the same weekend in 2017. Complete parts a through c. Click here to view the shopping data. Click here to view the t-distribution table data. a. Test the hypothesis that shoppers spent less than an average of $935.58 over the Thanksgiving weekend in 2017 using a = 0.05. Determine the null and alternative hypotheses. i Shopping Data Hoiu 2 935.58 H1:< 935.58 (Type integers or decimals. Do not round. Do not include the $ symbol in your answers.) $1,027 $871 $972 $959 $1,033 $989 $1,014 $1,027 $977 $836 $994 $880 $822 $921 $817 $661 Determine the appropriate critical value. Select the correct choice below and fill in the answer box within your choice. (Round to three decimal places as needed.) Print Done A. - ta = - 1.753 O B. ta/2= OC. ta = Calculate the appropriate test statistic. t = -0.41 (Round to two decimal places as needed.)

Answer 1

	X	x^2
1	1027	1054729
2	871	758641
3	972	944784
4	836	698896
5	822	675684
6	817	667489
7	959	919681
8	1033	1067089
9	989	978121
10	994	988036
11	921	848241
12	661	436921
13	1014	1028196
14	1027	1054729
15	977	954529
16	880	774400
Total	14800	13850166
Mean	925
SD	103.3331

Mean = $1588008119706_blob.png$

SD =

Σχ2 - (Σα)2 VV a η -1

Since we are testing whether the avg. spending is less than $935.58 .

Η :μ2 935.58

$H_{1}:\mu< 935.58$

This one tailed test since we are testing for left side.

Critical value with $1588008368987_blob.png$ = 0.05 = 5%

C.V. = $\mathbf{-t_{\alpha}}$

C.V. = 1.753 ......................using df = n-1 = 15 and p = 5%

Test Stat =

1 - Ho Sr/n

Where the null mean $1588008333247_blob.png$ = 935.58

Test Stat = -0.102

Since |test stat| < |C.V.|

We do not reject the null hypothesis. There is not sufficient evidence to conclude that shoppers spent less than an avg. of 935.58 over the Thanksgiving weekend in 2017.

b. the critical value at $\alpha$ = 10%

C.V. = 1.3406

We would still not reject at 10% so that would mean the p -value  > 10%.

C.V. = 0.866

We would still not reject at 20% so that would mean the p -value  > 20%.

Therefore

One tails: greater than 0.200

p-value = $P(t_{n-1}>|T.S.|)$

= $P(t_{15}>0.10)$

p -value = 0.4599 ....................using t-dist tables and excel 'tdist(|t.s.|df=15,tails=1)