How does the graph of an inverse function compare to the original function?

A market research firm knows from historical data that telephone surveys have a 36% response rate. In a random sample of 280 tel

Rasek [7]

Answer:

0.6604

Step-by-step explanation:

Given that a market research firm knows from historical data that telephone surveys have a 36% response rate.

Sample size of random sample = 280

We know for samples randomly drawn of large size sample proportion follows a normal distribution with mean= sample proportin and std error

= $\sqrt{\frac{pq}{n} }$

Substitute p = 0.36 and q = 1-0.36= 0.64

p follows N with mean = 0.36 and std dev = $\sqrt{\frac{0.36*0.64}{\sqrt{280} } } \\=0.0287$

Using normal distribution values we can find\

$P(33.5p.c. < p < 39pc)\\= P(0.335$

3 0

2 years ago

Read 2 more answers

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and

Brums [2.3K]

Answer:

$t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349$

$p_v =P(t_9>10.349)=1.34x10^{-6}$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=51.6$ represent the sample mean

$s=1.1$ represent the standard deviation for the sample

$n=10$ sample size

$\mu_o =48$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean score is higher than 48, the system of hypothesis would be:

Null hypothesis: $\mu \geq 48$

Alternative hypothesis: $\mu > 48$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349$

Calculate the P-value

First we find the degrees of freedom:

$df=n-1=10-1=9$

Since is a one-side upper test the p value would be:

$p_v =P(t_9>10.349)=1.34x10^{-6}$

Conclusion

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.

5 0

3 years ago

Solve the proportion.<br> 3/6 = 7/d

AysviL [449]

To solve this proportion:

$\frac{3}{6} =\frac{7}{d}$