Alabor economist is interested in the proportion of new ph. d.’s in economics who are female. he contacts a random sample of 100 new ph. d.’s in economics. of these, 40 are female. test the null hypothesis that the proportion of new ph. d.’s in economics who are female is at least 0.5, against the alternative hypothesis that it is less. use a 10% level of significance. give the value(s) of the critical value(s).

a. 0.5641
b. 0.4359
c. 0.4359 and 0.5641
d. - 1.82 and 1.82

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=40 represent the female new Ph.D.’s in economics

estimated proportion of female new Ph.D.’s in economics

is the value that we want to test

represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim given by: the null hypothesis that the proportion of new Ph.D.’s in economics who are female is at least 0.5, against the alternative hypothesis that it is les.:  

Null hypothesis:  

Alternative hypothesis:  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

(1)  

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

4) Critical value

For this case we can find the critical value with the normal standard distribution we need a value that accumulates 0.1 of the area on the left tail. And we can find it with the following excel code :"=NORM.INV(0.1;0;1)". And the critical value owuld be .

Now in order to find the critical value in terms of p we know this:

and replacing the values that we got :

And solving for we got:

2

step-by-step explanation:

quit

step-by-step explanation: