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.

Question

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

Answer 1

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Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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.