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Vesna [10]
2 years ago
8

The owner of Matt’s Gas Station wants to study gasoline purchases of customers at his station. In 2017, the mean amount of gas

oline purchased was 9 gallons. Matt wants to test whether or not the mean has changed from 9 gallons in 2018 to determine whether he needs to adjust how often gasoline deliveries are made to his station. The population standard deviation is known to be 2.5 gallons. To test this hypothesis, Horace takes a sample of 50 customer orders in 2018, and computes a sample mean of 9.9 gallons.
Required:
a. At the 0.01 level of significance, is there evidence that the population mean purchase is different from 9 gallons? Use the critical value approach to test this hypothesis and formally state your result.
b. Do we need to assume that the population of customer gasoline purchases is normally distributed? Explain.
c. If you were to construct a 99% confidence interval estimate for the population mean, would p = 9 be contained in the interval? Explain.
Mathematics
1 answer:
gulaghasi [49]2 years ago
4 0

Answer:

Step-by-step explanation:

a) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 9

For the alternative hypothesis,

µ ≠ 9

This is a 2 tailed test

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 9

x = 9.9

σ = 2.5

n = 50

z = (9.9 - 9)/(2.5/√50) = 2.55

Looking at the normal distribution table, the probability corresponding to the area above the z score is 1 - 0.99461 = 0.00539

Recall, population mean is 9

The difference between sample sample mean and population mean is 9.9 - 9 = 0.9

Since the curve is symmetrical and it is a two tailed test, the x value for the left tail is 9 - 0.9 = 8.1

the x value for the right tail is 9 + 0.9 = 9.9

These means are higher and lower than the null mean. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area. We already got the area above the z score as z = 0.00539

We would double this area to include the area in the left tail of z = - 2.55. Thus

p = 0.00539 × 2 = 0.01078

Since alpha, 0.01 < 0.01078, we would reject the null hypothesis

But we are told to use the critical value approach, then

Since α = 0.01, the critical value is determined from the normal distribution table.

For the left, α/2 = 0.01/2 = 0.005

The z score for an area to the left of 0.005 is - 2.575

For the right, α/2 = 1 - 0.005 = 0.995

The z score for an area to the right of 0.995 is 2.575

In order to reject the null hypothesis, the test statistic must be smaller than - 2.575 or greater than 2.575

Since - 2.55 > - 2.575 and 2.55 < 2.575, we would reject the null hypothesis. This corresponds to our previous decision.

b) we would assume normal distribution because the sample size is sufficiently large and the population standard deviation is known.

c) Confidence interval is written in the form,

(Sample mean ± margin of error)

Since the sample size is large and the population standard deviation is known, we would use the following formula and determine the z score from the normal distribution table.

Margin of error = z × σ/√n

The z score for confidence level of 99% is 2.58

Margin of error = 2.58 × 2.5/√50 = 0.91

Confidence interval = 9.9 ± 0.91

The lower end of the confidence interval is

9.9 - 0.91 = 8.99

Therefore, p = 9 will be contained in the interval.

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Answer:

t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784    

df=n-1=28-1=27  

p_v =P(t_{(27)}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=60 represent the sample mean

s=8.9 represent the sample standard deviation

n=28 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0.01 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 check if the true mean is less than 63, the system of hypothesis would be:  

Null hypothesis:\mu \geq 63  

Alternative hypothesis:\mu < 63  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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{s}{\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{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784    

P-value

The first step is calculate the degrees of freedom, on this case:  

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

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Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.  

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GrogVix [38]

<em>7 grams of sugar</em>

<h2>Explanation:</h2>

________________________________________

I'll assume your answer is as follows:

<em>A bowl of cereal had 3 3/6 grams of sugar in it. If Victor ate 2 bowl a week, how many grams of sugar would he have eaten?</em>

________________________________________

If so, then we know that:

  • A bowl of cereal had 3 3/6 grams of sugar in it.
  • Victor ate 2 bowl a week

In other words, the total of sugar he ate can be calculated as:

3\frac{3}{6}\times2 \\ \\ Converting \ mixed \ fraction \ into \ improper \ fraction: \\ \\ =(3+\frac{3}{6})\times2 \\ \\ =(\frac{3(6)+3}{6})\times 2 \\ \\ =(\frac{21}{6})\times 2 \\ \\ =(\frac{7}{2})\times 2 \\ \\ =7

So <em>he'd have eaten 7 grams of sugar</em>

<h2>Learn more:</h2>

Mixed numbers: brainly.com/question/383603

#LearnWithBrainly

7 0
3 years ago
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