Question

Question Hele ha survey of 1000 drivers from Region A 876 war a seat belt in a way of drivers from Region 3,923 war abelt 001, is there evidence to suport the came that the proportion des who wear seat belts in Region A is loss than the proportion of who were botsin Rajon? Asume that the samples are random and independent. Complete ports through below Ventily the daim Lal population 1 bedies from Region and let population be drivers from Region B. Choose the correct answer below. CA The comitato proportion des who were bills in Region Also than the proportion of new best in Region OB. The dair is that the proportion of hers who were boli Region Alsless than the proportion of drivers who were bots in Region OC. The aim is the proceso de who wear bronse OD The cams that the ofrivers who wants in both regions and Sund, Choose the correct answer bow OAH H₂. DISP2 OD 10:30 HD OB HP BE HD Ho Dip OCH H2 OR HER HD de the con ognised the core choice bow with complete your choice OA. There is one, ordre din regions Pound to decimals OB. There are till and two rejection regard OC. There is and creation > Find the dar Click to you 7 Type here to search 0 1 PER And Po 2 5 4 3 5 6 7 B LE 9 0 s O E R T Y U o Р Caps A S D G Н. K

Answer 1

For sample 1, we have that the sample size is Ni = 1000, the number of favorable cases is X1 = 870, so then the sample proportion is 1 = 870 = 0.87 1000 For sample 2, we have that the sample size is N2 = 1000, the number of favorable cases is X2 = 923, so then the sample proportion is p2 = x = 923 = 0.923 1000 The value of the pooled proportion is computed as p = x +X; 0.8965 870+923 1000+1000 Also, the given significance level is a = 0.01. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: P1 = P2 Ha: P1 P2 This corresponds to a left-tailed test, for which a z-test for two population proportions needs to be conducted. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the critical value for a left-tailed test is ze = -2.33. The rejection region for this left-tailed test is R= {z: z< -2.33} (3) Test Statistics The Z-statistic is computed as follows: Z P1 - 2 P(1 - 0) (1/nı + 1/n2) 0.87 – 0.923 0.8965. (1 - 0.8965) (1/1000 + 1/1000) = -3.891 (4) Decision about the null hypothesis Since it is observed that z = -3.891 < 2 = -2.33, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0001, and since p= 0.0001 <0.01, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population proportion p. is less than P2, at the 0.01 significance level. Confidence Interval The 99% confidence interval for pı - P2 is: -0.088 <P1 - P2 -0.018.

Claim is represented by 2nd option.

b. Option A is correct.

Critical Valie = -2. 33

Critical region = Z < -2.33

c. Test Statistics = -3. 89

d.

Since Z is falls into thr rejection region, Reject H0.

There is enough evidence to support the claim at 0.01.,.....less than......

Dear student,
I am waiting for your feedback. I have given my 100% to solve your queries. If you satisfied with my answer then please please like this.
Thank You