Answer:
Since the calculated value of z= 1.869 does not fall in the critical region we accept the null hypothesis H0: p1≠ p2 the data provides evidence that the proportion of high school students in metropolitan areas who have internet access during school hours is different than the proportion of rural high school students who have internet access during school hours.
Step-by-step explanation:
Let p1= proportion of the high school students having internet access
p2= proportion of the rural school students having internet access
1) The null and alternate hypothesis are
H0: p1≠ p2 against the claim Ha: p1= p2
2) We choose significance level ∝ =0.05
3) The test statistic under H0 is
z= p1^- p2^/√ p^q^( 1/n1 + 1/n2)
Now
p1^= 768/850= 0.9035
p2^= 308/355= 0.8670
p^= 768+ 308/850+ 355= 1076/1205
p^= 0.8929
q^= 1-p^= 0.1070
Putting the values
z= p1^- p2^/ √p^q^( 1/n1 + 1/n2)
Z= 0.9035-0.8670/sqrt [0.8929*0.1070( 1/850 + 1/355)]
z= 0.0365/ sqrt [ 0.0955403 (0.001176 + 0.002816)]
z= 0.0365/ 0.019531
z= 1.8688
The critical region is z∝/2 = ± 1.96
The value of z is 1.8666. The value of p is 0.06148 which is greater than 0.05
Conclusion:
Since the calculated value of z= 1.869 does not fall in the critical region we accept the null hypothesis H0: p1≠ p2 the data provides evidence that the proportion of high school students in metropolitan areas who have internet access during school hours is different than the proportion of rural high school students who have internet access during school hours.
