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cestrela7 [59]
3 years ago
10

ms turner drove 825 miles in march. she drove 3 times as many miles i march as she did in january.she drove 4 times as many mile

s in february as she did in january. what was the total number of miles ms tuer drove in february?
Mathematics
1 answer:
Natalija [7]3 years ago
8 0
Well first you have find out how many miles she drove in march. To find that you have to divide 825 by 3. That equals 275. Well if she drove 4 times more you take 275 x 4 and that equals 1100. She drove 1100 miles.
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Answer:

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So on this case the 95% confidence interval would be given by (25.06;33.74)  

b) z=\frac{29,4-25}{\frac{7}{\sqrt{10}}}=1.988  

p_v =P(z>1.988)=0.0234    

If we compare the p value and the significance level given \alpha=Step-by-step explanation:Previous conceptsA confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  
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Part a
Data given: 30 30 42 35 22 33 31 29 19 23
We can calculate the sample mean with the following formula:[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}= 29.4 the sample mean

\mu population mean (variable of interest)  

\sigma=7 represent the population standard deviation  

n=10 represent the sample size  

95% confidence interval  

The confidence interval for the mean is given by the following formula:  

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)  

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96  

Now we have everything in order to replace into formula (1):  

29.4-1.96\frac{7}{\sqrt{10}}=25.06  

29.4+1.96\frac{7}{\sqrt{10}}=33.74  

So on this case the 95% confidence interval would be given by (25.06;33.74)  

Part b

What are H0 and Ha for this study?  

Null hypothesis:  \mu \leq 25  

Alternative hypothesis :\mu>25  

Compute the test statistic

The statistic for this case is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-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".  

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

z=\frac{29,4-25}{\frac{7}{\sqrt{10}}}=1.988  

Give the appropriate conclusion for the test

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

p_v =P(z>1.988)=0.0234  

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.    

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