Question

For the hypotheses below test alpha = 0.025 with n = 100 and bar p = 0.67. State

Answer 1

Answer:

Part a : Critical value

For this case we want a value of $\alpha = 0.025$ we need to find on the normal standard distribution a z score that accumulates 0.025 of th area on the left and this value is $z_{crit}= -1.96$

Part b: Calculate the statistic  

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

$z=\frac{0.67 -0.76}{\sqrt{\frac{0.76(1-0.76)}{100}}}=-2.107$  

Part c: Statistical decision  

$n \hat p= 100*0.67= 67 >5$

$n (1-\hat p)= 100*(1-0.67)= 33 >5$

So then the normal approximation makes sense.

Since the calculated value is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level provided and we can conclude that the true proportion is lower than 0.76

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

$\hat p=0.67$ estimated proportion of interest

$p_o=0.76$ is the value that we want to test

$\alpha=0.025$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.76:  

Null hypothesis:$p\geq 0.76$  

Alternative hypothesis:$p < 0.76$  

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

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

Part a : Critical value

For this case we want a value of $\alpha = 0.025$ we need to find on the normal standard distribution a z score that accumulates 0.025 of th area on the left and this value is $z_{crit}= -1.96$

Part b: Calculate the statistic  

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

$z=\frac{0.67 -0.76}{\sqrt{\frac{0.76(1-0.76)}{100}}}=-2.107$  

Part c: Statistical decision  

$n \hat p= 100*0.67= 67 >5$

$n (1-\hat p)= 100*(1-0.67)= 33 >5$

So then the normal approximation makes sense.

Since the calculated value is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level provided and we can conclude that the true proportion is lower than 0.76