Statistics Solved Problem Samples

SAMPLE STATS PROBLEMS:
Problem 1: A researcher doing a fast food study claimed that the variance of the carbohydrate content of grilled chicken sandwiches at Arby's is higher than the variance of the carbohydrate content of grilled chicken sandwiches at McDonald's. The sample statistics are given below. Follow the guidelines on page 536 in our textbook, and test the researcher's claim using the Ftest with \(\alpha\) = 0.05.
Arby's: n_{1} = 16, \({{s}_{1}}^{2}\) = 4.84
McDonald's: n_{2} = 12, \({{s}_{2}}^{2}\) = 3.24
Solution: We need to test the following hypotheses
\[{{H}_{0}}:\sigma _{1}^{2}\le \sigma _{2}^{2}\]
\[{{H}_{A}}:\sigma _{1}^{2}>\sigma _{2}^{2}\]
We are going to use a righttailed F test. The critical value for 15, 11 degrees of freedom and significance level 0.05 is equal to \({{F}_{c}}=2.7186\). The Fstatistics is computed as
\[F=\frac{s_{1}^{2}}{s_{2}^{2}}=\frac{4.84}{3.24}=1.493827\]
Since \(F<{{F}_{c}}\), then we fail to reject the null hypothesis. That means that we don't have enough evidence to claim that variances are different, at the 0.05 significance level.
Problem 2: Experience raising New Jersey Red chickens revealed the mean weight of the chickens at five months is 4.35 pounds. The weights follow the normal distribution. In an effort to increase their weight, a special additive is added to the chicken feed. The subsequent weights of a sample of fivemonthold chickens were (in pounds):
4.41 4.37 4.33 4.35 4.30 4.39 4.36 4.38 4.40 4.39
At the .01 level, has the special additive increased the mean weight of the chickens? Estimate the pvalue.
Solution: We need to test the following hypotheses
\[{{H}_{0}}:\mu =4.35\]
\[{{H}_{A}}:\mu >4.35\]
Since the population variance is unknown, and the sample size is not big enough, we use a righttailed ttest. The tstatistics is computed as
\[t=\frac{\bar{X}\mu }{s/\sqrt{n}}\]
Using the sample we obtain that:
\[\bar{X}=4.368, s=0.033928\]
This means that
\[t=\frac{\bar{X}\mu }{s/\sqrt{n}}=\frac{4.3684.35}{0.033928/\sqrt{10}}=1.677698\]
The critical tvalue for this left tailed test for 9 degrees of freedom and \(\alpha =0.01\) is 2.821. Since the tstatistics is less than the tcritical, we fail reject the null hypothesis. In other words, we don't have enough evidence to claim that special additive increased the mean weight of the chickens, at the 0.01 significance level.
The pvalue is computed as
\[p=\Pr \left( t>1.677698 \right)=0.063859\]
Problem 3: A coin was flipped 60 times and came up heads 38 times.
(a) At the .10 level of significance, is the coin biased toward heads? Show your decision rule and calculations.
(b) Calculate a pvalue and interpret it.
Solution: (a) The following are the null and alternative hypotheses:
\[{{H}_{0}}:p\le 0.5\]
\[{{H}_{A}}:p> 0.5\]
Based on the above hypotheses, we need to use a righttailed ztest for proportions. We set the significance level at \(\alpha = 0.10\). The critical value is computed as
\[{{z}_{c}}=\text{1}\text{.28}\]
The rejection region for the null hypothesis is given by
\[z>{{z}_{c}}=\text{1}\text{.28}\]
Also, from the information provided we know that
\[\hat{p}=\frac{\text{38}}{\text{60}}=\text{0}\text{.633}\]
The zscore is computed as
\[z=\frac{\hat{p}p}{\sqrt{\frac{\left( 1p \right)}{n}}}=\frac{\text{0}\text{.633}\text{0}\text{.5}}{\sqrt{\frac{\text{0}\text{.5}\left( 1\text{0}\text{.5} \right)}{\text{60}}}}=\text{2}\text{.066}\]
The righttailed critical zvalue for \(\alpha = 0.10\) is given by
\[{{z}_{c}}=1.28\]
Since we have that \(z = 2.066 > {{z}_{c} }= 1.28\), then we reject the null hypothesis.
The righttailed pvalue is computed as
\[p=\Pr \left(z > 2.066 \right)=0.0194\]
(This means that the probability of getting a result as extreme as this assuming that the coin is fair is 0.0194).
Since we have that the pvalue is less than the significance level then we reject the null hypothesis. This implies that we have enough evidence to support the claim that the coin biased toward heads, at the 0.10 significance level.
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