205 206 Procter and Gamble (P&G;) wants to analyze the average weekly sales of t
ID: 3262393 • Letter: 2
Question
205 206 Procter and Gamble (P&G;) wants to analyze the average weekly sales of their Gilette Fusion razor in supermarkets that have a video advertisement for the razor a supermarkets that do not. Use the data in the Razor sheet in this workbook to generate e that have a video advertisement for the razor and in excel output to answer the following questions. 210 211 Quetion SA Can you conclude the average weekly sales of the Gillete Fusion razor in supermarkets that have a 214 that do not at a-0.17 State the bypethesis in terms of Video No Video. F video advertisement for the razor is greater than the average weekly sales in supermarkets Step I: Ha Step 3 ilin row 230 if the hypothess is two tailed. The smaller number must be typed fest Step 4: Step Sc Step 6: 24: Oaeste, sa Constructa86% confide ce interval for the difference i Population averages State theCL in terms of Video No Video. Lower Endpoint Upper Endpoint cel referencing descriptive statistics off the Escel outp a generated for Parts A and a (and using a looked up a table wala averages. Stabe the C.i., in terms of video No Video. Do NOT use the Confidence Interval section of the Excel output to selve t construct a 90% confidence nten al for the sme ence in th e a eston sc Lower Endpoint 4 HWNotes H HWExcelDirections Hw Camera Workforce Razor+Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: Video< No video
Alternative hypothesis: Video > No video
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 25.197
DF = 56
tcritical = 1.297
t = [ (x1 - x2) - d ] / SE
t = 2.98
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means (10) produced a t statistic of 2.98. We use the t Distribution Calculator to find P(t > 2.98) = 0.002133
Therefore, the P-value in this analysis is 0.002133.
Interpret results. Since the P-value (0.002133) is less than the significance level (0.10), we cannot reject the null hypothesis.
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