Listed below are the # of homeruns for the National League leader over the last 20 years. Assuming that # of homeruns is normally distributed, if this is sample data collected from a population of all past & future homerun leaders, test the claim that the mean homerun leader has less than 47 homeruns, where α=.05. Set up & complete the appropriate hypothesis test. For this data, also compute the p-value. Finally, compute 85% & 98% Confidence Intervals for this data.
Year (Homerun)
2006 (58)
2005 (51)
2004 (48)
2003 (47)
2002 (49)
2001 (73)
2000 (50)
1999 (65)
1998 (70)
1997 (49)
1996 (47)
1995 (40)
1994 (43)
1993 (46)
1992 (35)
1991 (38)
1990 (40)
1989 (47)
1988 (39)
1987 (49)
(The sum of all the homeruns is 984. The mean is 49.2)
I know this should be a left-tailed test since it states "less than". But I'm very confused about how to set it up. Would I state Ho:u=47 (null) & H1:u<47 (alt)?
If anyone can help (or at least point me in the right direction), I'd appreciate it.
Thanks!
To test the claim that the mean homerun leader has less than 47 homeruns, we can set up the following hypotheses:
Null Hypothesis (H0): The mean homerun leader is equal to 47.
Alternative Hypothesis (H1): The mean homerun leader is less than 47.
So yes, you are correct in stating Ho: μ=47 (null) and H1: μ<47 (alternative).
Next, we can perform a one-sample t-test to test these hypotheses. To do this, we need to calculate the test statistic and the p-value.
Step 1: Calculate the test statistic (t-value):
The test statistic for a one-sample t-test is calculated using the formula:
t = (x̄ - μ) / (s / √n)
Where x̄ is the sample mean, μ is the mean stated in the null hypothesis, s is the sample standard deviation, and n is the sample size.
In this case, the sample mean (x̄) is 49.2, the population mean (μ) stated in the null hypothesis is 47, the sample standard deviation (s) is unknown, and the sample size (n) is 20.
Since the population standard deviation is unknown, we can estimate it using the sample standard deviation formula:
s = √(Σ(x - x̄)² / (n - 1))
where Σ(x - x̄)² is the sum of squared differences between each data point and the sample mean, and n - 1 is the degrees of freedom.
Using the given data, we have:
Σ(x - x̄)² = (58 - 49.2)² + (51 - 49.2)² + ... + (49 - 49.2)²
= 305.76 + 2.56 + ... + 0.04 (calculated for each data point)
= 709.04
s = √(709.04 / 19)
≈ √37.32
≈ 6.106
Now we can calculate the t-value:
t = (49.2 - 47) / (6.106 / √20)
≈ 2.2
Step 2: Calculate the p-value:
To calculate the p-value, we need to find the probability of obtaining a more extreme test statistic (t-value) assuming the null hypothesis is true.
Since the alternative hypothesis is one-tailed (less than), we want to find the area under the t-distribution curve to the left of the t-value.
Using a t-distribution table or statistical software, we can find the p-value associated with the t-value of 2.2. Assuming a significance level (α) of 0.05, the p-value turns out to be less than 0.05 (significant).
Step 3: Interpret the results:
Since the p-value is less than our chosen significance level of 0.05, we reject the null hypothesis. This provides strong evidence to support the claim that the mean homerun leader has less than 47 homeruns.
Finally, to compute the confidence intervals, we can use the formulas for confidence intervals for a mean:
Confidence Interval = x̄ ± (t-value * (s / √n))
For an 85% confidence interval:
t-value = value from a t-distribution table with (n - 1) degrees of freedom and an alpha of (1 - 0.85) / 2
= value from a t-distribution table with 19 degrees of freedom and an alpha of 0.075
Using the appropriate t-value, we can calculate the lower and upper bounds for the 85% confidence interval:
Lower Bound = 49.2 - (t-value * (s / √n))
Upper Bound = 49.2 + (t-value * (s / √n))
Similarly, you can calculate the 98% confidence interval using the appropriate t-value for a 98% confidence level.
Note: If you want to compute the actual confidence intervals, you will need the critical t-values specific to the degrees of freedom for each confidence level mentioned above.