Question

Complete the following four hypotheses, using α = 0.05 for each.

a. Mean sales for all stores is at least $52,000 (Mean sales = $53,332.8, standard dev = 21,448.14, n = 100)

b. Proportion of stores with a medium market size is less than 55% (Medium market size n = 52, market size = 100)

c. Mean sales for promotion #3 is more than $45,000 per week (Mean sales for promo 3 = $61,899.38, n = 48, standard dev = 19,625.52)

d. Mean store age is greater than 8 years.

1. Using the same data set from part A, perform the hypothesis test for each speculation in order to see if there is evidence to support the manager's belief. Use the Seven Elements of a Test of Hypothesis from Section 7.1 of your text book, as well as the p-value calculation from Section 7.3, and explain your conclusion in simple terms.

2. Compute 99% confidence intervals for each of the variables described in a.-d., and interpret these intervals.

3. Write a report about the results, distilling down the results in a way that would be understandable to someone who does not know statistics. Clear explanations and interpretations are critical.

Answer 1

$1. a.~\mu=population~mean~sale\\ Null~hypothesis,~H_0:\mu=52000,~Alternative~hypothesis,~H_a:\mu<52000\\ Test~statistic=t=\frac{\sqrt{n}(\bar{x}-52000)}{s}=0.62\\ where,~\bar{x}=sample~mean=53332.8,~s=sample~sd=21448.14,~n=100\\ Critical~value=-t_{0.05,99}=-1.66\\ Since~Value~of~test~statistic=t>-1.66~and~\\ p-value=P(t<0.62|t\sim t_{99})=0.732>0.05\\ so~we~fail~to~reject~H_0~at~level~0.05~and~conclude~that~the~mean~sales\\~for~all~stores~is~at~least~\$52000.\\$

$2.a.~99\%~C.I.~for~\mu~is\\ (\$47700,~ \$58966)\\ 3.a.~Since~we~fail~to~reject~H_0~at~level~0.05~and~conclude~that~the~mean~sales\\~for~all~stores~is~at~least~\$52000.~Moreover~since~52000~falls~inside~the~99\%\\ ~C.I.~for~\mu,~we~are~99\%~confident~that~mean~sales~for~all~stores~is~at~least~\$52000.\\$

$1. b.~p=population~proportion\\ Null~hypothesis,~H_0:p=0.55,~Alternative~hypothesis,~H_a:p<0.55\\ Test~statistic=Z=\frac{\hat{p}-0.55}{\sqrt{\frac{0.55(1-0.55)}{100}}}=-0.60\\ where,~\hat{p}=sample~proportion=52/100=0.52\\ Critical~value=-Z_{0.05}=-1.64\\ Since~Value~of~test~statistic=Z>-1.64~and~\\ p-value=P(Z<-0.60|Z\sim N(0,1))=0.273>0.05\\ so~we~fail~to~reject~H_0~at~level~0.05~and~conclude~that~the~proportion\\ of~stores~with~a~medium~market~size~is~not~less~than~55\%.\\$

$2.a.~99\%~C.I.~for~p~is\\ (0.391312,~ 0.648688)\\ 3.a.~Since~~we~fail~to~reject~H_0~at~level~0.05~and~conclude~that~the~proportion\\ of~stores~with~a~medium~market~size~is~not~less~than~55\%.\\.$

$1. c.~\mu=population~mean~sale\\ Null~hypothesis,~H_0:\mu=45000,~Alternative~hypothesis,~H_a:\mu>45000\\ Test~statistic=t=\frac{\sqrt{n}(\bar{x}-45000)}{s}=5.97\\ where,~\bar{x}=sample~mean=61899.38,~s=sample~sd=19625.52,~n=48\\ Critical~value=t_{0.05,47}=1.68\\ Since~Value~of~test~statistic=t>1.68~and~\\ p-value=P(t>5.97|t\sim t_{47})=0.000<0.05\\ so~we~reject~H_0~at~level~0.05~and~conclude~that~the~mean~sales\\~for~all~stores~is~greater~than~\$45000.\\$

$2.c.~99\%~C.I.~for~\mu~is\\ (\$54295, ~\$69504)\\ 3.c.~Since~we~reject~H_0~at~level~0.05~and~conclude~that~the~mean~sales\\~for~all~stores~is~greater~than~\$45000.~Moreover~since~45000~is~less~than~the~lower~bound~of~the~99\%\\ ~C.I.~for~\mu,~we~are~99\%~confident~that~mean~sales~for~all~stores~is~greater~than~\$45000.\\$