In: Statistics and Probability
(Round all intermediate calculations to at least 4
decimal places.)
According to a Health of Boston report, female residents in Boston
have a higher average life expectancy as compared to male residents
(The Boston Globe, August 16, 2010). You collect the
following sample data to verify the results of the report. You also
use the historical (population) standard deviation of 8.2 years for
females and 8.6 years for males. (You may find it useful to
reference the appropriate table: z table
or t table)
Female | Male | ||||
x−1x−1 | = | 81.1 | x−2x−2 | = | 74.8 |
n1 | = | 32 | n2 | = | 32 |
Assume that μ1 is the population mean of life expectancy for females and μ2 is the population mean of life expectancy for males.
a. Set up the hypotheses to test whether the
average life expectancy of female Bostonians is higher than that of
male Bostonians.
H0: μ1 − μ2 = 0; HA: μ1 − μ2 ≠ 0
H0: μ1 − μ2 ≥ 0; HA: μ1 − μ2 < 0
H0: μ1 − μ2 ≤ 0; HA: μ1 − μ2 > 0
b. Calculate the value of the test statistic and
the p-value. (Round "Test statistic" value to 2
decimal places and "p-value" to 4 decimal
places.)
c. At the 10% significance level, what is the
conclusion?
Reject H0; females live longer than males.
Reject H0; females do not live longer than males.
Do not reject H0; females live longer than males
Do not reject H0; females do not live longer than males
a) H0: μ1 − μ2 ≤ 0; HA: μ1 − μ2 > 0
b)
sample #1 ------->
mean of sample 1, x̅1= 81.1000
population std dev of sample 1, σ1 =
8.2
size of sample 1, n1= 32
sample #2 --------->
1.023519452
mean of sample 2, x̅2= 74.8000
population std dev of sample 2, σ2 =
8.6
size of sample 2, n2= 32
difference in sample means = x̅1 - x̅2 =
81.1 - 74.8 =
6.3
std error , SE = √(σ1²/n1+σ2²/n2) =
2.1006
Z-statistic = ((x̅1 - x̅2)-µd)/SE =
6.3 / 2.1006 =
3.00
p-value =
0.0014 [excel function
=NORMSDIST(z)]
c) Reject H0; females live longer than males.