Which expression is equal to sina?

Mathematics

1 answer:

egoroff_w [7]2 years ago

6 0

Answer:

I'm not really sure with my answer,, but ai think it d heheheheh

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Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below.

garik1379 [7]

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.

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".

$\bar X_1 =9.2$ 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 $t_{\alpha/2}=1.96$