4 cm

A research group needs to determine a 90% confidence interval for the mean repair cost for all car insurance small claims. From

lutik1710 [3]

Answer:

a) $z = 1.645$

b) The should sample at least 293 small claims.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$ , which means that the answer of question a is z = 1.645.

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

(b) If the group wants their estimate to have a maximum error of $12, how many small claims should they sample?

They should sample at least n small claims, in which n is found when

$M = 12, \sigma = 124.88$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$12 = 1.645*\frac{124.88}{\sqrt{n}}$

$12\sqrt{n} = 205.43$

$\sqrt{n} = \frac{205.43}{12}$

$\sqrt{n} = 17.12$

$\sqrt{n}^{2} = (17.12)^{2}$

$n = 293$

The should sample at least 293 small claims.

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3 years ago

While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem

Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = population proportion.

So, Null Hypothesis, $H_0$ : p = 0.013 {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, $H_A$ : p > 0.013 {means that this is an unusually high number of faulty modems}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion faulty modems= $\frac{10}{367}$ = 0.027

n = sample of modems = 367

So, the test statistics = $\frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }$

= 2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that this is an unusually high number of faulty modems.

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4 years ago

Answerrrrrrr plsssssss

Genrish500 [490]

-8 i think? hope this helps!

6 0

3 years ago

(X*4^8)/4^5 Help I need help TONIGHT

mina [271]

x • 4^8 x
———— = ———— = x • 4^3 = x • 64 = 64x
4^5 4^(-3)

message me for a detailed explanation if you would like!

4 0