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jasenka [17]
2 years ago
11

A safety officer wants to prove that μ = the average speed of cars driven by a school is less than 25 mph. Suppose that a random

sample of 14 cars shows an average speed of 24.0 mph, with a sample standard deviation of 2.2 mph. Assume that the speeds of cars are normally distributed. For a significance level of α = 0.05, do we reject the null hypothesis? Which of the following is an appropriate conclusion?
Mathematics
1 answer:
melomori [17]2 years ago
7 0

Answer:

t=\frac{24-25}{\frac{2.2}{\sqrt{14}}}=-1.70    

p_v =P(t_{(13)}  

If we compare the p value and the significance level given \alpha=0.05 and we can 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 average speed is lower than 25 mph at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=24 represent the sample mean

s=2.2 represent the sample standard deviation

n=14 sample size  

\mu_o =25 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 check if the true mean for the average speed is lower than 25 mph, the system of hypothesis would be:  

Null hypothesis:\mu \geq 25  

Alternative hypothesis:\mu < 25  

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)  

Calculate the statistic

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

t=\frac{24-25}{\frac{2.2}{\sqrt{14}}}=-1.70    

P-value

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

df=n-1=14-1=13  

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

p_v =P(t_{(13)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 and we can 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 average speed is lower than 25 mph at 5% of signficance.  

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Answer:

2 cups of popcorn and 1/2 cups of cereal per cup

Step-by-step explanation:

khan told me

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2 years ago
On a coordinate plane, a piecewise function has 3 lines. The graph shows cleaning time in hours on the x-axis and total cost in
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Answer:

  C. (2, 6] hour jobs cost $100

Step-by-step explanation:

Let's consider each of these statements in view of the graph:

  1. A cleaning time of 2 hours will cost $100. -- The closed circle at (2, 50) tells you the cost of a 2-hour job is $50, not $100.
  2. A cleaning time of 6 hours will cost $150. -- The closed circle at (6, 100) tells you the cost of a 6-hour job is $100, not $150.
  3. Cost is a fixed rate of $100 for jobs requiring more than 2 hours, up to a maximum of 6 hours. -- The line between the open circle at (2, 100) and the closed circle at (6, 100) tells you this is TRUE.
  4. Cost is a fixed rate of $200 for jobs that require at least 6 hours. -- "At least 6 hours" means "greater than or equal to 6 hours." The closed circle at (6, 100) means a 6-hour job is $100, not $200.

3 0
2 years ago
Abdullah is a quality control expert at a factory that paints car parts. He knows that 20\ , percent of parts have an error in t
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The error rate has decreased after changing the painting process.

<u>Step-by-step explanation:</u>

Abdulla knows that 20 percent of the parts have an error in their painting. After suggesting changes in painting process, he wants to know whether the error rate has changed.

Number of parts in the random sample=400400400

Number of parts that had an error=606060

We have to determine what percentage of 400400400 is 606060

606060=x/100 \times 400400400\\=0.15%

After changing the painting process 0.15% of parts have error.

The previous percentage was 20.Hence the error rate has clearly changed.

3 0
2 years ago
7. Classical linear model assumptions for time series Consider the following stochastic process {(x1, x2, x3, . . . , xk, yt): t
iogann1982 [59]

Answer:

Options are missing.

The options for the above question are:

TS.1: Linear in parameters.

TS.2:No perfect collinearity

TS.3: Zero conditional mean.

TS.4: Homoskedasticity.

TS.5: No serial correlation

TS.6: Normality.

Hence the correct answer is TS1 to TS 5

Step-by-step explanation:

Assumptions TS 1 to TS 5 are the minimum set of assumptions needed to for the OLS estimates to be the best linear unbiased estimators conditional on explanatory variables for all time periods.

The assumptions of Normality is not needed for the estimators to show the BLUE property

3 0
2 years ago
A school cafeteria sells milk at 25 cents per carton and salads at 45 cents each. one week the total sales for these items were
denis-greek [22]

solution:

Lets start with the most amount that could have been sold.......using guess and check, we can figure out that 290 salads could have been sold, while 8 cartons of milk would have been sold.

The least amount of salads that could have been sold were none.

so,

you have  0<s<290

at least none were sold, and at most 290 were sold

but I do believe you are missing part of the question


4 0
2 years ago
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