**Educational Text Transcription**---A report just came out that stated the distribution of favorite ice cream flavors among Americans as follows: 32.5% vanilla, 20.5% chocolate, 17% butter pecan, 8% strawberry, and the rest other flavors. Cheryl wonders if her shop's customer preferences differ. The table below shows the results from 935 of her customers’ ice cream selections. What can be concluded at a 0.05 significance level?**Table: Frequencies of Favorite Ice Cream**| Outcome | Frequency | Expected Frequency ||-------------|-----------|--------------------|| Vanilla | 216 | 227 || Chocolate | 195 | 200 || Butter Pecan| 92 | 159 || Strawberry | 78 | 75 || Other | 354 | 263 |a. **Complete the table by filling in the expected frequencies** (Provided as above).b. **What is the correct statistical test to use?** - Answer: Goodness-of-Fit Testc. **What are the null and alternative hypotheses?** - \(H_0\): - The distribution of favorite ice cream for customers at her shop is the same as it is for Americans in general. - \(H_a\): - The distribution of favorite ice cream for customers at her shop is not the same as it is for Americans in general.d. **The degrees of freedom =** 4e. **The test statistic for this data =** \( \chi^2 = 84.08 \) (Approximate to two decimal places)f. **The p-value for this sample =** 0.0000 (Answer to four decimal places)g. **Based on this, we should:** - Reject \(H_0\)h. **Thus, the final conclusion is:** - There is sufficient evidence to conclude that the distribution of favorite ice cream for customers at her shop is not the same as it is for Americans in general.---**Diagrams and Explanations**There is a table that lists ice cream flavors with frequencies and expected frequencies. The test checks if Cheryl's shop distribution matches the national data using a Goodness-of-Fit test. The results show significant differences, leading to rejection of the null hypothesis about flavor preferences being identical in both

Solution: Given information: n= 955 Total observation. k= 5 categories α=0.05 Level of significance d) The degree of freedom is = df= k-1 df= 5-1 df= 4 e) Test statistic χ2=∑(O-E)2E Where O = observed frequency and E= Expected frequency Outcomes O E (O-E) (O-E)2 (O-E)2E Vanilia 216 227 -11 121 0.53304 Chocolate 195 200 -5 25 0.125 Butter Pocan 91 87 4 16 0.183908 Strawberry 92 78 14 196 2.512821 Other 361 363 -2 4 0.011019 ∑(O-E)2E= 3.365787χ2= 3.366 The test statistic for this data is 3.366 f) To find p value p value = P(χ2>3.366)p value = 0.4985 From Excel The p value for this sample is 0.4985 g) The p value is greater than α=0.05 h) Based on this we should fail to reject the null hypothesis i) The final conclusion is There is insuffficient evidence to conclude that the distribution of favoriate ice cream for customers at her shop is not same as it is for Americans in general. Option D is Correct