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.

Null hypothesis:  

Alternative hypothesis:  

And the best option for this case would be:

H0: p = 0.4 versus Ha: p < 0.4

So the p value obtained was a very high value and using the significance level given we have 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

estimated proportion of successes  

is the value that we want to test

represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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:  

Alternative hypothesis:  

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:

When we conduct a proportion test we need to use the z statisitc, 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 .

Calculate the statistic  

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

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 . The next step would be calculate the p value for this test.  

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

So the p value obtained was a very high value and using the significance level given we have 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.

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this

is there a picture

step-by-step explanation: