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trasher [3.6K]
3 years ago
14

Suppose that a manager is interested in estimating the average amount of money customers spend in her store. After sampling 36 t

ransactions at random, she found that the average amount spent was $ 35.25 35.25 . She then computed a 95 95 % confidence interval to be between $ 31.84 31.84 and $ 38.66 Which statement gives a valid interpretation of the interval? The store manager is 95% confident that the average amount spent by the 36 sampled customers is between S31.84 and $38.66. There is a 95% chance that the mean amount spent by all customers is between $31.84 and $38.66. O There is a 95% chance that a randomly selected customer will spend between $31.84 and $38.66. The store manager is 95% confident that the average amount spent by all customers is between $31.84 and S3866
Mathematics
1 answer:
Kitty [74]3 years ago
4 0

Answer: The managers statment would most likely give the valid interpretation.

Step-by-step explanation: If the store manager is 95% confident that the average amount spent was between $31.84 and $38.66. Then the store manager is most likey correct so I would have to say I agree with the manager. This may be a tricky question at first, but look at the problem from a different perspective. I am very sorry if this happens to be wrong.

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A random sample of n = 40 observations from a quantitative population produced a mean x = 2.2 and a standard deviation s = 0.29.
zmey [24]

Answer:

t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18    

df=n-1=40-1=39  

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

p_v =P(t_{(39)}>2.18)=0.0177  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and then the population mes seems to be higher than 2.1 at 5% of significance

Step-by-step explanation:

Data given and notation  

\bar X=2.2 represent the sample mean

s=0.29 represent the sample standard deviation

n=40 sample size  

\mu_o =2.1 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 mean is higher than 2,1, the system of hypothesis would be:  

Null hypothesis:\mu \leq 2.1  

Alternative hypothesis:\mu > 2.1  

If we analyze the size for the sample is > 30 but 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{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18    

P-value

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

df=n-1=40-1=39  

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

p_v =P(t_{(39)}>2.18)=0.0177  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and then the population mes seems to be higher than 2.1 at 5% of significance

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