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Masteriza [31]
3 years ago
14

The local swim team is considering offering a new semi-private class aimed at entry-level swimmers, but needs a minimum number o

f swimmers to sign up in order to be cost effective. Last year’s data showed that during eight swim sessions the average number of entry-level swimmers attending was 15. Suppose the instructor wants to conduct a hypothesis test and the alternative hypothesis is "the population mean is greater than 15." If the sample size is five, σ is known, and α = .01, the critical value of z is _______.
Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
7 0

Answer:

The significance level is \alpha=0.01 and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

z_{\alpha/2}= 2.326

So we reject the null hypothesis is z>2.326

Step-by-step explanation:

For this case we define the random variable X as the number of entry-level swimmers and we are interested about the true population mean for this variable . On specific we want to test this:

Null hypothesis: \mu \leq 15

Alternative hypothesis: \mu > 15

And the statistic is given by:

z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

The significance level is \alpha=0.01 and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

z_{\alpha/2}= 2.326

So we reject the null hypothesis is z>2.326

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Write this number in standard form. 300+80+0.9+0.06+0.001
Lina20 [59]

Hey there! I'm happy to help!

First, let's add the hundreds and the tens.

300+80=380

We see that there is nothing in the ones place, so we keep our ones place 0 and we move onto adding the tenths.

380+0.9=380.9

We add the hundredths.

380.9+0.06=380.96

And finally, we add the thousandths.

380.96+0.001=380.961

Therefore, this number in standard form is 380.961.

Have a wonderful day! :D

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25 divided by 2305 is 0.0106
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How do you round a decimal to the nearest hundredth.
Dafna11 [192]
Rounding decimals is very similar to rounding other numbers. If the thousandths place of a decimal is four or less, it is dropped and the hundredths place does not change. For example, rounding 0.843 to the nearest hundredth would give 0.84.

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3 years ago
The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation St
AysviL [449]

Answer:

a) 0.0668 = 6.68% probability that a domestic airfare is $550 or more

b) 0.1093 = 10.93% probability than a domestic airfare is $250 or less

c) 0.6313 = 63.13% probability that a domestic airfare is between $300 and $500

d) The cost for the 3% highest domestic airfares are $592 and higher.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 385, \sigma = 110

a. What is the probability that a domestic airfare is $550 or more (to 4 decimals)?

This is 1 subtracted by the pvalue of Z when X = 550. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{550 - 385}{110}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% probability that a domestic airfare is $550 or more

b. What is the probability than a domestic airfare is $250 or less (to 4 decimals)?

This is the pvalue of Z when X = 250. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{250 - 385}{110}

Z = -1.23

Z = -1.23 has a pvalue of 0.1093

0.1093 = 10.93% probability than a domestic airfare is $250 or less

c. What if the probability that a domestic airfare is between $300 and $500 (to 4 decimals)?

This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 300. So

X = 500

Z = \frac{X - \mu}{\sigma}

Z = \frac{500 - 385}{110}

Z = 1.045

Z = 1.045 has a pvalue of 0.8519

X = 300

Z = \frac{X - \mu}{\sigma}

Z = \frac{300 - 385}{110}

Z = -0.77

Z = -0.77 has a pvalue of 0.2206

0.8519 - 0.2206 = 0.6313

0.6313 = 63.13% probability that a domestic airfare is between $300 and $500

d. What is the cost for the 3% highest domestic airfares?

At least the 97th percentile, so at least X when Z has a pvalue of 0.97. So X when Z = 1.88.

Z = \frac{X - \mu}{\sigma}

1.88 = \frac{X - 385}{110}

X - 385 = 1.88*110

X = 592

The cost for the 3% highest domestic airfares are $592 and higher.

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