the mean weakly sales of soap bar in department stores was 146.3bars store.after an advertising campaign the mean weakly sales in 22 stores for a typical week increased to 153.7 and showed a S.D. 17.2 was the advertising campagin successful?
(t valve at 5% level of significance at (n-1)=21d.f=9.03)
Answers
Answer:
i think yhe q is wrong
Step-by-step explanation:
Answer:
campaign was failure
Step-by-step explanation:
The provided sample mean is \bar{x}=153.7xˉ=153.7 and the sample standard deviation is s=17.2,s=17.2, and the sample size is n=22.n=22.
The following null and alternative hypotheses need to be tested:
H_0: \mu=146.3H0:μ=146.3
H_1: \mu>146.3H1:μ>146.3
This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.
Based on the information provided, the significance level is \alpha=0.05, df=n-1=22-1=21α=0.05,df=n−1=22−1=21 and the critical value for a right-tailed test is t_c=1.721tc=1.721
The rejection region for this right-tailed test is R=\{t:t>1.721\}.R={t:t>1.721}.
The t-statistic is computed as follows:
t={\bar{x}-\mu_0 \over s/\sqrt{n}}={153.7-146.3 \over 17.2/\sqrt{22}}\approx2.018t=s/nxˉ−μ0=17.2/22153.7−146.3≈2.018
Since it is observed that t=2.018>1.7221=t_c,t=2.018>1.7221=tc, then it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean is greater than 146.3, at the 0.05 significance level.
Using the P-value approach:
t=2.018, df=22-1, \alpha=0.05, one-tailedt=2.018,df=22−1,α=0.05,one−tailed
The p-value is p=0.028274,p=0.028274, and since p=0.028274<0.05=\alpha,p=0.028274<0.05=α, it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean is greater than 146.3, at the 0.05 significance level.
Hence the advertising campaign was successful in promoting sales.