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avanturin [10]
3 years ago
10

A college entrance exam company determined that a score of 2222 on the mathematics portion of the exam suggests that a student i

s ready for​ college-level mathematics. To achieve this​ goal, the company recommends that students take a core curriculum of math courses in high school. Suppose a random sample of 150150 students who completed this core set of courses results in a mean math score of 22.722.7 on the college entrance exam with a standard deviation of 3.93.9. Do these results suggest that students who complete the core curriculum are ready for​ college-level mathematics? That​ is, are they scoring above 2222 on the math portion of the​ exam? Complete parts​ a) through​ d) below.
Mathematics
1 answer:
kolbaska11 [484]3 years ago
7 0

Answer:

Yes, Students are scoring above 2222 on the math portion of the​ exam.

Step-by-step explanation:

We are given that a college entrance exam company determined that a score of 2222 on the mathematics portion of the exam suggests that a student is ready for​ college-level mathematics i.e., population mean is 22.

Null Hypothesis, H_0 : \mu = 22 {Students are scoring equal to 2222 on the math portion of the​ exam}

Alternate Hypothesis, H_0 : \mu > 22 {Students are scoring above 2222 on the math portion of the​ exam}

Also, a random sample of 150 students who completed this core set of courses results in a mean math score of 22.7 on the college entrance exam with a standard deviation of 3.93 i.e.;

Sample mean, Xbar = 22.7    and Sample standard deviation, s = 3.93

The Test statistics is given by;

                            Z = \frac{Xbar -\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

Test Statistics = \frac{22.7 -22}{\frac{3.93}{\sqrt{150} } } ~ t_1_4_9

                        = 2.1815

Since, we are not given with the significance level, so we assume it to be 5%. At 5% significance level t table gives critical value of 1.6578 at 149 degree of freedom. Since our test statistics is more than the critical value so we reject null hypothesis as our test statistics fall in the rejection region.

Therefore, we conclude that Students are scoring above 2222 on the math portion of the​ exam.

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a) z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236  

p_v =2*P(Z>0.236)=0.813  

If we compare the p value and using any significance level for example \alpha=0.01 always p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant differences between the two proportions.  

b) We are confident at 99% that the difference between the two proportions is between -0.188 \leq p_B -p_A \leq 0.226

Step-by-step explanation:

Previous concepts and data given

A 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".  

p_A represent the real population proportion of women age 18 to 30  agreed with the statement that a woman should have the right to a legal abortion for any reason

\hat p_A =\frac{26}{75}=0.347 represent the estimated proportion of women age 18 to 30  agreed with the statement that a woman should have the right to a legal abortion for any reason

n_A=75 is the sample size for A

p_B represent the real population proportion for women age 58 to 70  agreed with the statement that a woman should have the right to a legal abortion for any reason

\hat p_B =\frac{21}{64}=0.328 represent the estimated proportion of women age 58 to 70  agreed with the statement that a woman should have the right to a legal abortion for any reason

n_B=64 is the sample size required for B

z represent the critical value for the margin of error and for the statisitc

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Part a

We need to conduct a hypothesis in order to check if the proportion are equal, the system of hypothesis would be:  

Null hypothesis:p_{A} = p_{B}  

Alternative hypothesis:p_{A} \neq p_{B}  

We need to apply a z test to compare proportions, and the statistic is given by:  

z=\frac{p_{A}-p_{B}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{A}}+\frac{1}{n_{B}})}}   (1)

Where \hat p=\frac{X_{A}+X_{B}}{n_{A}+n_{B}}=\frac{26+21}{75+64}=0.338

Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236  

Statistical decision

Since is a two sided test the p value would be:  

p_v =2*P(Z>0.236)=0.813  

If we compare the p value and using any significance level for example \alpha=0.01 always p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant differences between the two proportions.  

Part b  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.58  

And replacing into the confidence interval formula we got:  

(0.347-0.328) - 2.58 \sqrt{\frac{0.347(1-0.347)}{75} +\frac{0.328(1-0.328)}{64}}=-0.188  

(0.347-0.328) + 2.58 \sqrt{\frac{0.347(1-0.347)}{75} +\frac{0.328(1-0.328)}{64}}=0.226  

And the 99% confidence interval for the difference of proportions would be given (-0.188;0.226).  

We are confident at 99% that the difference between the two proportions is between -0.188 \leq p_B -p_A \leq 0.226

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