A random sample of n = 1,000 observations from a binomial population contained 385 successes. You wish to show that p < 0.4. State the null and alternative hypothesis. H0: p < 0.4 versus Ha: p > 0.4 H0: p ≠ 0.4 versus Ha: p = 0.4 H0: p = 0.4 versus Ha: p < 0.4 H0: p = 0.4 versus Ha: p > 0.4 H0: p = 0.4 versus Ha: p ≠ 0.4 Calculate the appropriate test statistic. (Round your answer to two decimal places.) z = Provide an α = 0.05 rejection region. (Round your answer to two decimal places. If the test is one-tailed, enter NONE for the unused region.) z > z < State your conclusion. H0 is not rejected. There is insufficient evidence to indicate that p is less than 0.4. H0 is rejected. There is insufficient evidence to indicate that p is less than 0.4. H0 is not rejected. There is sufficient evidence to indicate that p is less than 0.4. H0 is rejected. There is sufficient evidence to indicate that p is less than 0.4.

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion is not significantly lower than 0.4

H0 is not rejected. There is insufficient evidence to indicate that p is less than 0.4.

Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

X=385 represent the successes

[tex]\hat p=\frac{385}{1000}=0.385[/tex] estimated proportion of successes

[tex]p_o=0.4[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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 lower than 0.4.:

Null hypothesis:[tex]p\geq 0.4[/tex]

Alternative hypothesis:[tex]p < 0.4[/tex]

And the best option for this case would be:

H0: p = 0.4 versus Ha: p < 0.4

For this case we want a rejection region at 5% of significance, since we w are conducting a left tailed test we need to find a quantile on the normal standard distribution who accumulates 0.05 of the area on the left and we got: [tex]z_{cric}= -1.64[/tex]

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

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)

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

Calculate the statistic

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

[tex]z=\frac{0.385 -0.4}{\sqrt{\frac{0.4(1-0.4)}{1000}}}=-0.968[/tex]

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

[tex]p_v =P(z<-0.968)=0.1665[/tex]

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion is not significantly lower than 0.4

H0 is not rejected. There is insufficient evidence to indicate that p is less than 0.4.