Listed below are the number of homeruns for the National League leader over the last 20 years, through 2006. Assuming that number of homeruns is normally distributed, if this is sample data collected from a population of all past and future homerun leaders, test the claim that the mean homerun leader has less than 47 homeruns, where α=.05. Set up and complete the appropriate hypothesis test. For this data, also compute the p-value and describe how you could have used this information to complete the analysis. Finally, compute 85% and 98% Confidence Intervals for this data.
Year Homeruns Year Homeruns
2006 58 1996 47
2005 51 1995 40
2004 48 1994 43
2003 47 1993 46
2002 49 1992 35
2001 73 1991 38
2000 50 1990 40
1999 65 1989 47
1998 70 1988 39
1997 49 1987 49
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Listed below are the number of homeruns for the National League leader over the last 20 years, through 2006. Assuming that number of homeruns is normally distributed, if this is sample data collected from a population of all past and future homerun leaders, test the claim that the mean homerun leader has less than 47 homeruns, where α=.05. Set up and complete the appropriate hypothesis test. For this data, also compute the p-value and describe how you could have used this information to complete the analysis. Finally, compute 85% and 98% Confidence Intervals for this data.
Year Homeruns Year Homeruns
2006 58 1996 47
2005 51 1995 40
2004 48 1994 43
2003 47 1993 46
2002 49 1992 35
2001 73 1991 38
2000 50 1990 40
1999 65 1989 47
1998 70 1988 39
1997 49 1987 49
Year Homeruns Year Homeruns
2006 58 1996 47
2005 51 1995 40
2004 48 1994 43
2003 47 1993 46
2002 49 1992 35
2001 73 1991 38
2000 50 1990 40
1999 65 1989 47
1998 70 1988 39
1997 49 1987 49
DATA:
year ---- homeruns ----year ----homerus
2006 ----58 ------ 1996 ----47
2005 ----51 ------ 1995 ----40
2004 ----48 ------ 1994 ----43
2003 ----47 ------ 1993 ----46
2002 ----49 ------ 1992 ----35
2001 ----73 ------ 1991 ----38
2000 ----50 ------ 1990 ----40
1999 ----65 ------ 1989 ----47
1998 ----70 ------ 1988 ----39
1997 ----49 ------ 1987 ----49
Since a claim is being made that the mean homerun leader has less than 47 homeruns, this claim becomes the alternative hypothesis. The null hypothesis is what we suspect isn't true, while the alternative hypothesis is what we suspect is true (or claim to be true). The null hypothesis ALWAYS uses an equals sign. Therefore, you can set up this way:
Ho: µ ≥ 47 -->meaning the population mean is greater than or equal to 47.
Ha: µ < 47 -->meaning the population mean is less than 47.
Since it is assumed that the homeruns are normally distributed, you can probably use a one-sample z-test for the data. You will need to find the mean and standard deviation for this sample in order to be able to plug the values into the formula to find the z-statistic.
Here is the formula for a one-sample z-test:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
Use 47 for the population mean. Sample size is 20.
If the z-statistic exceeds the critical value from a z-table (find α=.05 for a one-tailed test), the null is rejected in favor of the alternative hypothesis and µ < 47 (there is enough evidence to support the claim). The test is one-tailed because the alternative hypothesis is showing a specific direction. The test is two-tailed when the alternative hypothesis says something like "does not equal" and doesn't specify a specific direction.
The p-value is the actual level of the test statistic and can be found using a z-table.
To find confidence intervals, find a z-value for those intervals and use a confidence interval formula like the following:
CI = mean + or - (z-value)(sd divided by √n)
...where + or - (z-value) represents the confidence interval using a z-table, sd = standard deviation, and n = sample size.
I hope this will help get you started.
To test the claim that the mean homerun leader has less than 47 homeruns, we need to perform a hypothesis test. Here is how you can set up and complete the appropriate hypothesis test:
1. State the null and alternative hypotheses:
- Null hypothesis (H₀): The mean homerun leader is equal to or greater than 47 homeruns.
- Alternative hypothesis (H₁): The mean homerun leader is less than 47 homeruns.
2. Determine the significance level: The given significance level (α) is 0.05.
3. Calculate the test statistic: We will use a one-sample t-test because the population standard deviation is unknown. The formula for the test statistic is:
t = (sample mean - hypothesized mean) / (sample standard deviation / √sample size)
4. Find the critical value(s) or p-value: Since we are performing a lower-tailed test, we need to find the critical t-value from the t-distribution based on the degrees of freedom (n - 1).
5. Make a decision: Compare the calculated t-value from step 3 with the critical t-value from step 4. If the calculated t-value is less than the critical t-value, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.
Now let's calculate the test statistic and make the decision:
Sample Mean (x̄) = sum of homeruns / number of years = (58 + 51 + ... + 49) / 20 ≈ 51.9
Sample Standard Deviation (s) = square root of [ (Sum of (homeruns - x̄)²) / (n - 1) ] = square root of [ ( (58-51.9)² + (51-51.9)² + ... + (49-51.9)² ) / 19 ] ≈ 9.85
t = (51.9 - 47) / (9.85 / √20) ≈ 1.63
Degrees of Freedom (df) = 20 - 1 = 19
Critical t-value: Using a t-table or a t-distribution calculator with α=0.05 and df=19, the critical t-value is approximately -1.729.
Since the calculated t-value (1.63) is greater than the critical t-value (-1.729), we fail to reject the null hypothesis.
Next, let's compute the p-value. The p-value is the probability of obtaining a test statistic as extreme as (or more extreme than) the observed value, assuming the null hypothesis is true.
To compute the p-value, we can use a t-distribution calculator with the calculated t-value and the degrees of freedom (df=19). The p-value for t=1.63 is approximately 0.059.
Since the p-value (0.059) is greater than the chosen significance level (0.05), we fail to reject the null hypothesis.
Therefore, based on the given sample data, we do not have enough evidence to support the claim that the mean homerun leader has fewer than 47 homeruns.
Now let's compute the 85% and 98% Confidence Intervals for this data:
For an 85% Confidence Interval:
- Calculate the margin of error: The margin of error is equal to the critical value (obtained from the t-distribution) multiplied by the standard error. The formula for the margin of error is: Margin of Error = critical value * (sample standard deviation / √sample size)
- Compute the lower and upper bounds: The lower bound is calculated by subtracting the margin of error from the sample mean, and the upper bound is calculated by adding the margin of error to the sample mean.
For a 98% Confidence Interval:
- Follow the same steps as for the 85% Confidence Interval, but use the appropriate critical value for a 98% level of confidence.
Using a t-table or a t-distribution calculator, you can obtain the critical values for the specified levels of confidence and degrees of freedom.
Note: The critical values will depend on the sample size and the chosen significance level.