Question

1 4 9 An important source of income for a large international banking company is home loans. The loan manager at a branch keeps track of the numbe accumulated the data shown in the table below. From previous years, the manager believes that the distribution of the number of customers ha using a significance level of 0.05 Number of 0 2 3 4 5 26 Customers Frequencies 1 11 14 7 10 State the appropriate null and alternative hypotheses. Choose the correct answer below. O A. HoThe population is Poisson distributed with mean equal to 3.7 HA The population is not Poisson distributed with mean equal to 3.7 O B. Ho: The population is not Poisson distributed with mean equal to 3.7 HA The population is Poisson distributed with mean equal to 3.7 OC. Ho: The population mean is not equal to 3.7 HA The population mean is equal to 3.7 OD. Ho The population mean is equal to 3.7 HA: The population mean is not equal to 3.7 The test statistic is y=0 Round to three decimal places as needed) The p-value is Round to three decimal places as needed.) Draw a conclusion. Choose the correct answer below. A. Reject the null hypothesis, and conclude that the population is not Poisson distributed with mean equal to 3.7 B. Do not reject the null hypothesis, and conclude that the population is Poisson distributed with mean equal to 3.7 C. Reject the null hypothesis, and conclude that the population mean is not equal to 3.7 D. Do not reject the null hypothesis, and conclude that the population mean is equal to 3.7 Il auto.proctoru.com is sharing yours

Answer 1

Here the hypothesis are

H₀ : The Population is poisson distributed with mean equal to 3.7.

H_a : The Population is not poisson distributed with mean equal to 3.7.

Here Total values = 56

Expected value = POISSON (x, 3.7; false) * 56

So here expected value and chi-square probability value

Number of customers	Frequencies	probability	Expected Value	Chi-square
0	1	0.0247	1.3845	0.1068
1	4	0.0915	5.1227	0.2461
2	9	0.1692	9.4770	0.0240
3	11	0.2087	11.6883	0.0405
4	14	0.1931	10.8117	0.9402
5	7	0.1429	8.0007	0.1252
>=6	10	0.1699	9.5151	0.0247
Sum	56	1	56	1.5075

$\chi^2$ = 1.507

Degree of freedom =dF = 7 -1 = 6

Here the p value = CHIDIST(1.507; 6) = 0.9590

Here we failed to reject the null hypothesis, and conclude that the population is Poisson distributed with mean 3.7.

Option B is correct here.