91影视

Consider X_M = 79 and n_M=2,447.

The redemption proportion for the milk-shake m-voucher test is,

$\overline{P{炉}_M}=\frac{x_M}{n_M}$

$=\frac{79}{2447}$

$=0.032$

Thus, the redemption rate for the sample of the milk-shake m-voucher is 0.032.

Consider x_D = 72 and n_D = 6,619.

The redemption rate for the sample of donut m-voucher is,

$$\begin{gathered} {\overline{P}}_M=\frac{x_D}{n_D} \\ =\frac{72}{6619} \end{gathered}$$

= 0.011

So, the reclamation rate for the sample of the donut m-voucher is 0.011.

$\overline{P}{}_M-\overline{P}{}_D=0.032-0.011$

= 0.021

Thus, the point estimate for the difference between the actual redemption rates is 0.021.

The critical value for a two-tailed test is obtained below:

Here, the test is two-tailed, and the significance level is $伪=0.10$

The rejection region for the two-tailed test is$\left|z_c\right|>z_{\frac{a}{2}}$.

The confidence coefficient is 0.90.

So,

(1-伪) = 0.90

$伪$= 0.10

$\frac{伪}{2}$= 0.05

From Appendix D Table II, the critical value for the two-tailed test with 伪 = 0.10 is 伪 =0.10 is $z_{\frac{a}{2}\left(-0.05\right)}$$z_{\frac{a}{2}\left(-0.05\right)}$= 卤1.645. so, the refusal region is $\left|z_c\right|$ > 1.645.

The 90% confidence interval is obtained below:

$\left[\frac{({\stackrel{炉}{P}}_M鈭�{\stackrel{炉}{P}}_D)卤z_{0.05}}{\sqrt{\frac{{\stackrel{炉}{P}}_M(1鈭�{\stackrel{炉}{P}}_M)}{n_M}+\frac{{\stackrel{炉}{P}}_D(1鈭�{\stackrel{炉}{P}}_0)}{n_D}}}\right]=0.021卤1.645\sqrt{\frac{0.032(1鈭�0.032)}{2447}+\frac{0.011(1鈭�0.011)}{6619}}$

$$\begin{gathered} =0.021卤1.645(0.00378) \\ =0.021卤0.006 \\ =(0.015,0.027) \end{gathered}$$

Thus, the 90% confidence interval is (0.015, 0.027).

Interpretation:

There is 90% confidence that the difference between the actual redemption rates between milk-shake m-voucher and donut m-voucher lies between 0.015 and 0.027.

Explanation:

The 90% of all similarly generated confidence intervals will contain the true value of the difference in redemption rates in repeated sampling.

Rule:

If the hypothesized value lies outside the corresponding 100(1-伪) % confidence interval, then reject the null hypothesis.

Here, the 90% confidence interval is (0.015. 0.827), which does not contain the hypothesized value of 0. The hypothesized value 0 lies outside the interval (0.015, 0.027). Thus, it can be concluded that the null hypothesis H₀ is rejected at = 0.10.

Hence, there is a statistically significant difference between the redemption rates.

If the genuine disparity among redeeming rates is more than 0.01, the disparity is considered "practically important."

Thus, there鈥檚 a "practically significant difference between the reclamation rates because the hypothesized value 0.01 lies outside the interval (0.015, 0.027).