Question #350105

1. the recommended daily calorie intake for teenage girls is 2,200 calories/day. a nutritionist at a state university believes the average daily caloric intake of girls in that state to be lower. test that hypothesis, at the 5% level of significance, against the null hypothesis that the population average is 2,200 calories/day using the following sample data: n=36,x¯=2,150,s=203.

1
Expert's answer
2022-06-13T14:58:30-0400

The following null and alternative hypotheses need to be tested:

H0:μ2200H_0:\mu\ge2200

H1:μ<2200H_1:\mu<2200

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,\alpha = 0.05, df=n1=35df=n-1=35 and the critical value for a left-tailed test is tc=2.030108.t_c =- 2.030108.

The rejection region for this left-tailed test is R={t:t<2.030108}.R = \{t:t<- 2.030108\}.

The t-statistic is computed as follows:


t=xˉμs/n=21502200203/36=1.4778t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{2150-2200}{203/\sqrt{36}}=-1.4778


Since it is observed that t=1.4778>2.030108=tc,t=-1.4778>- 2.030108=t_c, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed, df=35df=35 degrees of freedom, t=1.4778t=-1.4778 is p=0.074202,p=0.074202, and since p=0.074202>0.05=α,p=0.074202>0.05=\alpha, it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean μ\mu

is less than 2200, at the α=0.05\alpha = 0.05 significance level.


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