Mathematics, 23.12.2019 18:31 AnonymousLizard52303

Apolling organization reported data from a survey of 2000 randomly selected canadians who carry debit cards. participants in this survey were asked what they considered the minimum purchase amount for which it would be acceptable to use a debit card. suppose that the sample mean and standard deviation were $9.11 and $7.30, respectively. (these values are consistent with a histogram of the sample data that appears in the report.) do these data provide convincing evidence that the mean minimum purchase amount for which canadians consider the use of a debit card to be appropriate is less than $10? carry out a hypothesis test with a significance level of 0.01. (use a statistical computer package to calculate the p-value. round your test statistic to two decimal places and your p-value to three decimal places.) t = ? p-value = ?

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Answer from: gwbdee918

Null hypothesis: $\mu \geq 10$

Alternative hypothesis : $\mu$

$t=\frac{9.11-10}{\frac{7.30}{\sqrt{2000}}}=-5.45$

$p_v =P(t_{(1999)}$

And we can use the following excel code: "=T.DIST(-5.45,1999,TRUE)"

Step-by-step explanation:

Data given and notation

$\bar X=9.11$ represent the sample mean

s=7.30 represent the sample standard deviation

n=2000 sample size

$\mu_o =5.70$ 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.

Is a one tailed left test.

What are H0 and Ha for this study?

Null hypothesis: $\mu \geq 10$

Alternative hypothesis : $\mu$

Find the rejection region for the test using significance level=.01

First we need to find the degrees of freedom on this case:

df=n-1=2000-1=1999

In order to find the critical value we need to find a value on the t distirbution such that:

$P(t_{(1999)}$ where 0.01 represent the significance level selected.

And this value is $t_{crit}=-2.33$

We can use the following excel code: "=T.INV(0.01,1999)"

The rejection regin would be (-infinity;-2.33)

Compute the test statistic

The statistic for this case 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:

$t=\frac{9.11-10}{\frac{7.30}{\sqrt{2000}}}=-5.45$

Give the appropriate conclusion for the test

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

$p_v =P(t_{(1999)}$

And we can use the following excel code: "=T.DIST(-5.45,1999,TRUE)"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

R code

Answer from: Quest

Okay so for every triangle has three angles that always add up to 180 degrees. and to tell the difference between whether they are secure, right or obtuse is that acute: has one angle that is less than 90 degrees. right triangle means there is at least one angle that is exactly 90 degrees. and an obtuse angle has an angle that is bigger than 180 degrees. i’ll do #13 as an example: 13. 40 and 50 so if every triangle has 180 degrees 40+50= 90 degrees then180-90= 90 degrees so the third angle is 90 degrees and because this triangle has a 90 degree angle it means it is a right triangle

Answer from: Quest

idk

step-by-step explanation:

i don't know