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jarptica [38.1K]
2 years ago
8

Can someone please help me with this :( Thank u :))

Mathematics
1 answer:
faltersainse [42]2 years ago
3 0

Answer:

It would be 54/100

Step-by-step explanation:

because the 4 in 5.4 is in the hundreds so its 54/100

pls mark as brainliest

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Answer:

The 95% confidence interval would be given by 0.281 \leq \mu_1 -\mu_2 \leq 0.529  

Step-by-step explanation:

Previous concepts  

A 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".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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\bar X_1 =9.2 represent the sample mean 1

\bar X_2 =8.8 represent the sample mean 2

n1=27 represent the sample 1 size  

n2=30 represent the sample 2 size  

\sigma_1 =0.3 population standard deviation for sample 1

\sigma_2 =0.1 population standard deviation for sample 2

\mu_1 -\mu_2 parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:

\bar X_1 -\bar X_2 =9.2-8.8=0.4

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 t_{\alpha/2}=1.96  

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

0.4-1.96\sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}=0.281  

0.4+1.96\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=0.519  

So on this case the 95% confidence interval would be given by 0.281 \leq \mu_1 -\mu_2 \leq 0.529  

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