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Somebody please answer those two I NEED THEM LIKE NOWW

DochEvi [55]

Answer:

already put answer lol i just want points

Step-by-step explanation:

6 0

3 years ago

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A random sample of size n1= 25, taken from a normal population with a standard deviation σ1= 5, has a mean X1= 80. A second rand

sesenic [268]

Answer:

The 94% confidence interval would be given by $2.898 \leq \mu_1 -\mu_2 \leq 7.102$

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.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =80$ represent the sample mean 1

$\bar X_2 =75$ represent the sample mean 2

n1=25 represent the sample 1 size

n2=36 represent the sample 2 size

$\sigma_1 =5$ sample standard deviation for sample 1

$\sigma_2 =3$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

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:

$\bar X_1 -\bar X_2 =80-75=5$

Since the Confidence is 0.94 or 94%, the value of $\alpha=0.06$ and $\alpha/2 =0.03$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that $z_{\alpha/2}=1.88$