In: Statistics and Probability
In a recent Gallup poll, 507 adults age 18 and older were
surveyed and 269 said they believed they would not have enough
money to live comfortably in retirement. Use a 0.01 significance
level to test the claim that the proportion of adults who do not
believe they will have enough money to live comfortably in
retirement is smaller than 60%.
The test statistic is:
[ Select ]
["-1.61", "-2.73", "-4.21", "-3.55", "-3.19"]
The p-value is: (to 4 decimals)
[ Select ]
["0.0007", "0.0001", "0.0032", "0.0002",
"0.0024"]
Based on this we:
[ Select ]
["Reject
the null hypothesis", "Fail to reject the null hypothesis"]
Conclusion There
[ Select ]
["does not", "does"]
appear to be enough evidence
to support the claim that the proportion of adults who do not
believe they will have enough money to live comfortably in
retirement is smaller than 90%.
Solution:
Given in the question
The claim is that the proportion of adults who do not believe they
will have enough money to live comfortably in retirement is smaller
than 60%. So null and alternate hypothesis can be written as
Null hypothesis H0: p = 0.6
Alternate hypothesis Ha: p <0.60
Number of sample n = 507
Number of favourable cases X = 269
Sample proportion(p^) = 269/507 = 0.53
Here we will use one proportion Z test because
np = 0.6*507 = 304.2
nq = 0.4*507 = 202.8
Both np and nq both are greater than 5 so test stat can be
calculated as
Z-test stat = (p^ - p)/sqrt(p*(1-p)/n) =
(0.53-0.6)/sqrt(0.6*(1-0.6)/507) = -0.07/0.0218 = -3.19
As this is left tailed test so from Z table we found p-value =
0.0007
At significance level
= 0.01, we can reject the null hypothesis as p-value is less than
alpha value (0.0007<0.01).
So we can say that there does appear to be enough evidence to
support the claim that the proportion of adults who do not believe
they will have enough money to live comfortably in retirement is
smaller than 60%.