Estimate the unit rate justify your answer $299 for 4 tires

A sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's p

Grace [21]

Answer:

Yes, we have sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

We are given that a sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 29% do not fail in the first 1000 hours of their use.

Let Null Hypothesis, $H_0$ : p $\leq$ 0.29 {means that less than or equal to 29% do not fail in the first 1000 hours of their use}

Alternate Hypothesis, $H_1$ : p > 0.29 {means that more than 29% do not fail in the first 1000 hours of their use}

The test statics that will be used here is One-sample proportions test;

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

where, $\hat p$ = proportion of chips that do not fail in the first 1000 hours of their use = 32%

n = sample of chips = 1500

So, <u>test statistics</u> = $\frac{0.32-0.29}{\sqrt{\frac{0.32(1-0.32)}{1500} } }$

= 2.491

<em>Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.</em>

Therefore, we conclude that more than 29% do not fail in the first 1000 hours of their use which means we have sufficient evidence at the 0.02 level to support the company's claim.

7 0

3 years ago

A rectangle is four times as long as it is wide. If it has an area of 36 square inches, what are its dimension?.

vivado [14]

Answer:

b on edge

Step-by-step explanation:

3 0

3 years ago

Assume that the playbook contains 10 passing plays and 12 running plays. The coach randomly selects 9 plays from the playbook. W

Nezavi [6.7K]

Answer:

6

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A new phone-answering system installed by a certain utility company is capable of handling twelve calls every 5 minutes. Prior

SVEN [57.7K]

Answer:

A. P(x>12 in 5 minutes)=0.0201

Step-by-step explanation:

Because we are working with a Poisson Distribution of probability, we need to get all the data needed. In a Poisson distribution is needed a constant called λ that symbolizes the mean data (6 calls) per unit of time (5 minutes), for this distribution λ=6/5.

Poisson probabilities work like this:

$P(x=y\ 'in\ z\ unit\ of\ time')= \frac{e^{-z\lambda}(z\lambda)^{y}}{y!}$

Remember y has to be an integer and the units of z must be the same unit of time used in λ. Now we are ready to solve this problem

A. The question is asking for the probability that in 5 minutes appear more calls than the phone-answering machine could answer (i.e P(x>12 in 5 minutes)). Because there are infinite numbers greater than 12, we are using this property of probabilities that´ll help us simplify the problem:

P(x>12 in 5 minutes)= 1 - P(x≤12 in 5 minutes)

Now we can use the following:

$P(x\geq12\ in\ 5\ minutes)=P(x=0\ in\ 5\ min)+P(x=1\ in\ 5\ min)+...P(x=12\ in\ 5\ min)$

$P(x\geq12\ in\ 5\ minutes)=\sum_{n=0}^{\infty}P(x=n\ in\ 5\min)$

$P(x\geq12\ in\ 5\ minutes)=\sum_{n=0}^{\infty}\frac{e^{-6}6^{n}}{n!}$

And you can find P(x=n in 5 minutes) for every n using a calculator or a computer, and finally add them to get

P(x≤12 in 5 minutes)= 0.9799

P(x>12 in 5 minutes)= 1-0.9799

This will be our answer

P(x>12 in 5 minutes)= 0.0201

8 0

3 years ago

ryan deposits $775 in an account that pays 1.24% simple interest for four years. Brian deposits $775 in an account thats 1.24% s

AleksAgata [21]

Answer:

Principal (p) = $775

Rate of interest (r) = 4.24% = 0.0424

No of years/time (t) = 4 years

Simple interest = ptr

= 775 × 4 × 0.0424

= 131.44 dollars

Hence simple interest after 4 years is $131.44

Amount = simple interest + principal

= 775 + 131.44

= 906.44 dollars

Hence amount after 4 years is $906.44

RESULT

906.44 dollars

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