Question

According to the U.S. Census Bureau, 11% of children in the United States lived with at least one grandparent in 2009 (USA TODAY, June 30, 2011). Suppose that in a recent sample of 1560 children, 249 were found to be living with at least one grandparent. At a 1% significance level, can you conclude that the proportion of all children in the United States who currently live with at least one grandparent is higher than 11%? Use both the p-value and the critical-value approaches. Round your answers for the observed value of z and the critical value of z to two decimal places, and the p-value to four decimal places.

Round your answers for the observed value of z and the critical value of z to two decimal places, and the p-value to four decimal places.

z_observed =_____

p-value =______

Critical value =_____

Answer 1

According to the U.S. Census Bureau, 11% of children in the United States lived with at least one grandparent in 2009 (USA TODAY, June 30, 2011). Suppose that in a recent sample of 1560 children, 249 were found to be living with at least one grandparent.

Now at 1% significance level we want to test that, the proportion of all children in the United States who currently live with at least one grandparent is higher than 11%.

[ 1% significance level means, ]

Null and Alternative Hypotheses:-

[ Where p = population proportion of all children in the United States who currently live with at least one grandparent ]

Here sample size n = 1560 i.e we can use a large sample test.

Now our appropriate test statistic is given by,

[ where, n = number of random sample = 1560, Sample proportion of success = (249/1560) = 0.1596, Hypothesized proportion = 0.11 ]

Value of test statistic:- ( putting the value of n, , ]

[ Round to two decimal places ]

Answer:- Observed_Z = 6.26 [ Round to two decimal places ]

• Critical Value Of Z:-

   [ value are getting from standard normal probability table and round to two decimal places ]

Answer:- Critical Value = 2.33 [ Round to two decimal places ]

• p-value of the test:-

p-value for right tailed test is given by,

[ where z = observed value of Z = 6.26 ]

[ We know that if Z~N(0,1) then , ]   

  [ From standard normal probability table we get, ]

Answer:- p-value = 0.00

Rejection Rule Based On Critical Value:-

We reject the null hypothesis at level   if we get, Observed_Z > Critical Value

Here we get, Observed_Z = 6.26 and Critical Value = 2.33

i.e we get, Observed_Z = 6.26 > Critical Value = 2.33

Result: We reject the null hypothesis.

Rejection Rule Based On p-value:-

We reject the null hypothesis at level if we get,

Here we get, p-value = 0.00 and

i.e we get, p-value = 0.00

Result: We reject the null hypothesis.

Conclusion:-

From the above testing result at a 1% significance level , we can conclude that, the proportion of all children in the United States who currently live with at least one grandparent is higher than 11% .