Write the standard form 12 ten thousands 8 thousands 14 hundreds 7 ones

Suppose that the researchers wanted to estimate the mean reaction time to within 6 msec with 95% confidence. Using the sample st

Lisa [10]

Answer:

a) The 95% confidence interval would be given by (509.592;550.308)

b) n=523 rounded up to the nearest integer

Step-by-step explanation:

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".

$\bar X=530$ represent the sample mean for the sample

$\mu$ population mean

s=70 represent the sample standard deviation

n=48 represent the sample size (variable of interest)

Confidence =95% or 0.995

Part a

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom are df=n-1=48-1=47

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: "=-T.INV(0.025,47)".And we see that $z_{\alpha/2}=2.01$

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

$530-2.01\frac{70}{\sqrt{48}}=509.692$

$530+2.01\frac{70}{\sqrt{48}}=550.308$

So on this case the 95% confidence interval would be given by (509.592;550.308)

Part b

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Assuming that $\hat \sigma =s$

And on this case we have that ME =6msec, and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval is provided, $z_{\alpha/2}=1.96$ , replacing into formula (b) we got:

$n=(\frac{1.96(70)}{6})^2 =522.88 \approx 523$

So the answer for this case would be n=523 rounded up to the nearest integer

3 0

2 years ago

The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in n

devlian [24]

Answer:

The length of the longest 15% of Atlantic cod in this area is 53.79cm, roundeed to 2 decimal places.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.

In this problem, we have that:

A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm, so $\mu = 49.9, \sigma = 3.74$ .

What is the length in cm of the longest 15% of Atlantic cod in this area?

We have to find the value of X for the value of Z that has a pvalue of 0.85.

Looking at the zscore table, we have that Z = 1.04 has a pvalue of 0.8508. So, we have to find the value of X when $Z = 1.04, \mu = 49.9, \sigma = 3.74$ .