Suppose a new standardized test is given to 100 randomly selected third- grade students in New Jersey. The sample average score Y on the test is 58 points, and the sample standard deviation, sy, is 8 points. a. The authors plan to administer the test to all third-grade students in New Jersey. Construct a 95% confidence interval for the mean score of all New Jersey third graders b. Suppose the same test is given to 200 randomly selected third graders from Iowa, producing a sample average of 62 po

1 answer
Question:

Suppose a new standardized test is given to 100 randomly selected third- grade students in New Jersey. The sample average score Y on the test is 58 points, and the sample standard deviation, sy, is 8 points.
a. The authors plan to administer the test to all third-grade students in New Jersey. Construct a 95% confidence interval for the mean score of all New Jersey third graders
b. Suppose the same test is given to 200 randomly selected third graders from Iowa, producing a sample average of 62 points and sample standard deviation of 11 points. Construct a 90% confidence interval for the difference in mean scores between Iowa and New Jersey.
c. Can you conclude with a high degree of confidence that the population means for Iowa and New Jersey students are different? (What is the standard error of the difference in the two sample means? What is the p-value of the test of no difference in means versus some difference?)

Answers

Answer:a) [tex]58-1.984\frac{8}{\sqrt{100}}=56.41[/tex]    [tex]58+1.984\frac{8}{\sqrt{100}}=59.59[/tex]    So on this case the 95% confidence interval would be given by (56.41;59.59)    b) [tex]62-1.653\frac{11}{\sqrt{200}}=60.71[/tex]    [tex]62+1.653\frac{11}{\sqrt{200}}=63.29[/tex]    So on this case the 90% confidence interval would be given by (60.71;63.29)c)  [tex]df=n_{1}+n_{2}-2=100+200-2=298[/tex][tex]p_v =2*P(t_{(298)}<-3.21)=0.0014[/tex]So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesisStep-by-step explanation:Previous conceptsA confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".The margin of error is the range of values below and above the sample statistic in a confidence interval.Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".[tex]\bar X[/tex] represent the sample mean for the sample  [tex]\mu[/tex] population mean (variable of interest)s represent the sample standard deviationn represent the sample size  Part aThe confidence interval for the mean is given by the following formula:[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:[tex]df=n-1=100-1=99[/tex]Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,99)".And we see that [tex]t_{\alpha/2}=1.984[/tex]Now we have everything in order to replace into formula (1):[tex]58-1.984\frac{8}{\sqrt{100}}=56.41[/tex]    [tex]58+1.984\frac{8}{\sqrt{100}}=59.59[/tex]    So on this case the 95% confidence interval would be given by (56.41;59.59)    Part bThe confidence interval for the mean is given by the following formula:   (1)In order to calculate the critical value we need to find first the degrees of freedom, given by:[tex]df=n-1=200-1=199[/tex]Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that [tex]t_{\alpha/2}=1.653[/tex]Now we have everything in order to replace into formula (1):[tex]62-1.653\frac{11}{\sqrt{200}}=60.71[/tex]    [tex]62+1.653\frac{11}{\sqrt{200}}=63.29[/tex]    So on this case the 90% confidence interval would be given by (60.71;63.29)Part cFor this case the two standard errors are given:[tex] SE= 1.984\frac{8}{\sqrt{100}}=1.59[/tex]    [tex] SE= 1.653\frac{11}{\sqrt{200}}=1.29[/tex]    [tex]\bar X_1=58[/tex] represent the mean for New Jersey[tex]\bar X_2=62[/tex] represent the mean for Iowa[tex]s_1=8[/tex] represent the sample standard deviation for New Jersey[tex]s_2=11[/tex] represent the sample standard deviation for Iowa[tex]n_1=100[/tex] sample size for New Jersey[tex]n_2=200[/tex] sample size for Iowat would represent the statistic (variable of interest)Concepts and formulas to useWe need to conduct a hypothesis in order to check if the mean are equal , the system of hypothesis would be:Null hypothesis:[tex]\mu_{1} = \mu_{2}[/tex]Alternative hypothesis:[tex]\mu_{1} \neq \mu_{2}[/tex]t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.Calculate the statisticWe can replace in the t formula the result obtained is:[tex]t=\frac{58-62}{\sqrt{\frac{(8)^2}{100}+\frac{(11)^2}{200}}}}=-3.21[/tex]  Statistical decisionFor this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:[tex]df=n_{1}+n_{2}-2=100+200-2=298[/tex]Since is a bilateral test the p value would be:[tex]p_v =2*P(t_{(298)}<-3.21)=0.0014[/tex]So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.

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