Question

* Im not actually sure if i got these right but really not sure about the bottom questions

gender selection, the proportion of baby girls is less than 0.5. 1 standard deviation below 0. Complete parts (a) through (h) A 0.01 significance level is used for a hypothesis test of the claim that when parents use partic ular method Assume that sample data consists of 36 girls in 81 births, so the sample statistic of a below Click here to view page 1 of the Normal table. Click here to view.age 2 of the Normal table a. Identify the null hypothesis and the alternative hypothesis. Choose the correct answer 28 OA. Ho: P05 HpX0.5 III OC. Ho p0.5 H:p>0.5 b. What is the value of a? 0.05 (Type an integer or a decimal.) c. What is the sampling distribution of the sample statistic? O O - Student (t) distribution * Normal distribution d. is the test two-tailed, lehtailed, or right-tailed? O Lof-taled Two-tailed Right-tailed O What is the value of the test statistic?

Answer 1

Part a)

Part b)

α = 0.01

Part c)

Normal distribution

Part d)

Left-tailed

Part e)

P = X / n = 36/81 = 0.4444

Test Statistic :-
Z = ( P - P0) / √(P0 * q0 / n)
Z = ( 0.4444 - 0.5 ) / √(( 0.5 * 0.5) /81)
Z = -1

Part f)

P value = P ( Z < -1 ) = 0.1587
Looking for the value Z = -1 in standard normal table to find the P value.

Part g)

Critical value

Z(α) = Z(0.01) = 2.33
Looking for the value 1 - α = 1 - 0.01 = 0.99 in standard normal table to find the critical value Z

Reject null hypothesis if Z < -Z(α)

Hence the critical value is -Z(α) = - Z(0.01) = - 2.33

Part h)

Critical region

Reject null hypothesis if Z < -Z(α) i.e  

Reject null hypothesis if Z < - 2.33