Question

2. In a three-digit lottery, each of the three digits is supposed to have the same probability of occurrence (counting initial blanks as zeros, e.g., 32 is treated as 032). The table shows the frequency of occurrence of each digit for 90 consecutive daily three-digit drawings. At a- 05, can you reject the hypothesis that the digits are from a uniform population? The summary data are shown below and may be found in the Excel file, Lottery3. Digit Frequency 0 33 1 17 2 25 3 30 4 31 5 28 6 24 7 25 8 32 9 25 Total 270 At the 0.05 level of significance, determine if the digits are from a uniform population State your hypotheses and show all 7 steps clearly. Be sure to write a conclusion that clearly and completely addresses the research question. Give and interpret the p-value. Attach or include your Minitab output.

a. STATE HYPOTHESIS and show all 7 steps clearly

b. GIVE AND INTERPRET P-VALUE

c. ATTACH MINITAB output, if you don't intend to attach minitab, please don't answer the question!!

Answer 1

Answer Date: 20/03/2018 Step 1 To test the hypothesis is that the digits are not uniform distribution at 5% level of significance. Step 2 The null and alternative hypothesis is, The null hypothesis is that the digits are uniform distribution The alternative hypothesis is that the digits are not uniform distribution. step 3 The chi-square test statistic for goodnessof-fit tst is, Byusing MINITAB software, find chi-square with help of following steps: 1) Import the data.

2) Select Chi-Square Goodness-of-Fit Test (One Variable) from Tables in Stat menu 3) Select Observed counts and Equal proportions 4) Click Ok Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: Observed Contribution to Chi-Sq 1.33333 3.70370 0.14815 0.33333 0.59259 0.03704 0.33333 0.14815 0.92593 0.14815 Test Category Observed Proportion Expected 17 25 30 31 28 24 25 32 25 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 27 27 27 27 27 27 27 27 27 27 9 10 N DFChi-Sq P-Value 0.564 270 9 7.70370 From the MINITAB output, the chi-square test statistic for goodness-of-fit test is 7.70 Step 4

The degree of freedom is, From the MINITAB output, the degree of freedom is 9 Step 5 The p-value is From the MINITAB output,the chi-squartestp-value is 0 564. The p-value is 0.564 Step 6 Decision The conclusion is that the significant p-value is greater than 0.05 which is 0.564, so the null hypothesis is not rejected at 5% level of significance. Step 7

There is insufficient evidence to indicate that the digits are not uniform distribution. The result is not statistically significant