A fruit juice franchise company has a policy of opening new fruit juice stand only on those areas that have a mean household income of at least ₱30,500 a month. The company is currently considering an area in which to open a new fruit juice stand. The company’s research department took a sample of 25 households from this area and found that the mean monthly income of these households is ₱32,600. Using 5% significance level, would you conclude that the company should open a fruit juice stand in the area? Also, find the 95% confidence interval of the true mean.
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Expert's answer
2022-05-05T04:15:09-0400
Let standard deviation is s=4500.
We need to construct the 95% confidence interval for the population mean μ.
The critical value for α=0.05 and df=n−1=24 degrees of freedom is tc=z1−α/2;n−1=2.063899.
The corresponding confidence interval is computed as shown below:
CI=(xˉ−tc×ns,xˉ+tc×ns)
=(32600−2.063899×254500,
32600+2.063899×254500)
=(30742.4909,34457.5091)
Therefore, based on the data provided, the 95% confidence interval for the population mean is 30742.4909<μ<34457.5091, which indicates that we are 95% confident that the true population mean μ is contained by the interval (30742.4909,34457.5091).
The following null and alternative hypotheses need to be tested:
H0:μ≥30500
Ha:μ<30500
This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.
Based on the information provided, the significance level is α=0.05,df=n−1=24 degrees of freedom, and the critical value for a left-tailed test is tc=−1.710882.
The rejection region for this left-tailed test is R={t:t<−1.710882}.
The t-statistic is computed as follows:
t=s/nxˉ−μ=4500/2532600−30500≈2.3333
Since it is observed that t=2.3333≥−1.711=tc, it is then concluded that the null hypothesis is not rejected.
Using the P-value approach:
The p-value for left-tailed test, df=24 degrees of freedom, t=2.3333 is p=0.98583, and since p=0.98583≥0.05=α, it is concluded that the null hypothesis is not rejected.
Therefore, there is not enough evidence to claim that the population mean μ
is less than 30500, at the α=0.05 significance level.
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