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Statistics - Baseball Team Scores

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Perform each of the steps below using Excel. Use the collected data from the chart (MLB Statistics) as a sample of the population (Any one statistic you choose). Describe the population that the data represents. Interpret each of your results with a conclusion and decision. Use any statistic on the MLB chart to answer the questions.

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This solution is provided in a .doc and .xls file attached. It calculates all the required statistics and analyzes them as required.

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See the attachment please.

1. Develop a 90% confidence interval and a 95% confidence interval for the population mean. Did you use a z statistic or t statistics? Why? Compare the size and meaning of the two confidence intervals.

We use the column-E: R. By using Excel, we get the following descriptive statistics
Column-E

Mean 281.1667
Standard Error 4.803276
Median 282.5

Mode 285
Standard Deviation 26.30862
Sample Variance 692.1437
Kurtosis -0.29514
Skewness 0.128215
Range 112
Minimum 222
Maximum 334
Sum 8435
Count 30

From above table, we know that the sample size n=30, sample mean =281.1667 and sample standard deviation s=26.30862. Use a website http://duke.usask.ca/~rbaker/Tables.html , we get .
So, a 90% confidence interval and a 95% confidence interval for the population mean are

and

i.e.,

and

We use t-statistics since we don't know the standard deviation of population. A 95% Confidence interval has a larger length than a 90% CI does. It tells us that the population mean lies in with probability 90%, and it lies in with 95% probability.

2. Test a hypothesis about a population mean. Take a random sample and perform a t-test against a population mean. Explain what assumptions you made.

We use the column-E: R. By using Excel, we get the following descriptive statistics
Column-E

Mean 281.1667
Standard Error 4.803276
Median 282.5

Mode 285
Standard Deviation 26.30862
Sample Variance 692.1437
Kurtosis -0.29514
Skewness 0.128215
Range 112
Minimum 222
Maximum 334
Sum 8435
Count 30

WE try to test the following hypotheses. Denote the population mean by .

From above table, we know that the sample size n=30, sample mean =281.1667 and sample standard deviation s=26.30862. By question 1), we know that a 95% confidence interval is . Now 270 is NOT in
So, with 5% significance level, we should reject the null hypothesis We conclude that

3. Test a hypothesis comparing two population means. Collect two independent samples. Perform a t-test using two independent samples.

We use the column-U and V: HBB-IBB. Use Excel, we get the following

t-Test: Two-Sample Assuming Equal Variances

Variable HBB Variable IBB
Mean 22.1 15.5
Variance 36.57586 26.32759
Observations 30 30
Pooled Variance 31.45172
Hypothesized Mean Difference 0
df 58
t Stat 4.557927
P(T<=t) one-tail 1.36E-05
t Critical one-tail 1.671553
P(T<=t) two-tail 2.71E-05
t Critical two-tail 2.001717

From the above table, we know that p-value=P(T<=t) two-tail=2.71E-05<0.05. So, with 5% significance level, we should reject the null hypothesis: No difference between their means. We conclude that there is a significant difference between their means.

4. Test a hypothesis comparing two population means using paired (matched) observations. Collect two dependent samples. Perform a t test for paired observations.

We use the column-U and V: HBB-IBB. Now we assume that two samples are dependent. Use Excel, we get the following

t-Test: Paired Two Sample for Means

Variable HBB Variable IBB
Mean 22.1 15.5
Variance 36.57586 26.32759
Observations 30 30
Pearson Correlation -0.08723
Hypothesized Mean Difference 0
df 29
t Stat 4.373603
P(T<=t) one-tail 7.2E-05
t Critical one-tail 1.699127
P(T<=t) two-tail 0.000144
t Critical two-tail 2.04523

From the above table, we know that ...

Solution provided by:
Education
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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