Question

the month of October, as shown in the table to the right Use a 0.01 sign A police department released the numbers of calls for the dferent days of the week during t with this analysis? Wec 243 Tues 227 Mon 200 Sun 152 Day Frequency Determine the null and alternative hypotheses. Ho H, Calculate the test statistic Round to three decimal places as needed) Calculate the P-value P-value (Round to four decimal places as needed) What is the conclusion for this hypothesis test? ○ A. Fal to reject Ho There is insufficient evidence to warrant rejection of the dam that the different days ofthe week have the same frequences of police cats O B. Fail to reject Ho ○ C. Reject Ho There is insumoent evidence to warrant rejection orme calm that the different days of the week have the same frequencies of police calls There is sufficient evidence to warrant rejection of the claim that the different days of the week have the same trequencies of police calls D, Reject Ho There is sumcent ovidence to warrant rejection of the claim that the different days of the week have the same tequences of police calls What is the fundamentai error with this analysis O A. Because October has 31 days, two of the days of the week occur more often than the other days of the week O B. Because October has 31 days, each day of the week occurs the same number of times as the other days of the week О с. Because October has 31 days, one othe days of the week occur more often than the other days ofthe week O D. Because October has 31 days, three of the days of the week occur more often than the other days of the week

sun 152

mon 200

Tues 227

wed 243

Thurs 177

Fri 208

sat 238

Answer 1

(fo-fe)2fe Categories Observed Expected (152-206.635)-/206.635 152 1445 0.143-206.635 sun 14.446 (200-206.635)-/206.635 200 1445*0.143-206.635 mon 0.213 (227-206.635)-/206.635 Tues 227 1445 0.143-206.635 2.007 (243-206.635)/206.635 6.4 (177-206.635)-/206.635 243 1445*0.143-206.635 wed Thurs 1445 0.143-206.635 4.25 (208-206.635)-/206.635 Fri 208 1445*0.143-206.635 0.009 (238-206.635)-/206.635 4.761 sat 238 1445 0.143-206.635 1445 32.086 Sum 1445 The following null and alternative hypotheses need to be tested Ho pi 0.143, p20.143, p3 0.143, p 0.143,p 0.143, p6 0.143,p 0.143 Ha: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit.

(2) Rejection Region Based on the information provided, the significance level is a 0.01, the number of degrees of freedom is df 7-1 6, so then the rejection region for this test is Rxx16.812) (3) Test Statistics The Chi-Squared statistic is computed as follows: 14.446 +0.213+2.007+6.4+4.25+0.009+ 4.761 32.086 (4)_Decision about the null hypothesis Since it is observed that x rejected. 32.08616.812, it is then concluded that the null hypothesis is It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the a- 0.01 significance level.

H₀ : Police calls occur with the same frequency on different days of the week

H₁ : Atleast one day has a different frequency of calls than the other days

232.086

P-value = 0

Conclusion :

Correct option : D. Reject H₀ .There is sufficient evidence to warrant the rejection of the claim that the different days of the week have the same frequency of police calls.

Fundamental error with analysis :

Correct option : D. Because the October has 31 days, three of the days of the week occur more often than the other days of the week.