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A random sample of 25 statistics examinations was selected. The average score in the sample was 76 with a variance of 144. Assum

Mazyrski [523]

Answer:

99% confidence interval for the population average examination score is between a lower limit of 69.2872 and an upper limit of 82.7128.

Step-by-step explanation:

Confidence interval = mean +/- margin of error (E)

mean = 76

variance = 144

sd = sqrt(variance) = sqrt(144) = 12

n = 25

degree of freedom (df) = n - 1 = 25 - 1 = 24

confidence level (C) = 99% = 0.99

significance level = 1 - C = 1 - 0.99 = 0.01 = 1%

t-value corresponding to 24 df and 1% significance level is 2.797

E = t×sd/√n = 2.797×12/√25 = 6.7128

Lower limit = mean - E = 76 - 6.7128 = 69.2872

Upper limit = mean + E = 76 + 6.7128 = 82.7128

99% confidence interval is (69.2872, 82.7128)

6 0

2 years ago

Dean needs $150 for a miming class. He already has $85 and he can earn the rest by working 13 hours at his job. If h represents

AnnyKZ [126]

Answer:

h=$65/13

Step-by-step explanation:

So he has $85

Now the remaining amount he needs is $65

And in every 13h he produces the $65

So to find his hourly earning is h=$65/13

So the equation is

h = $65/13

3 0

3 years ago

Which civilization believed in one god? a : babylonians b : egyptians c : hebrews d : phoenicians

Radda [10]

The Hebrews were the only civilization who believed one God.

5 0

2 years ago

Read 2 more answers

Just this last one left

Vlada [557]

Answer:

2240 scooter

Step-by-step explanation:

725+579+696=2000

2000*1.12=2240

7 0

2 years ago

Read 2 more answers

At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul

vampirchik [111]

Answer:

a) Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b) The 95% confidence interval would be given by (910.05;959.95)

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) $z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750$

Since is a bilateral test the p value is given by:

$p_v =2*P(Z>2.750)=0.0059$

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b. What is the 95% confidence interval estimate of the population mean examination score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

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".

$\bar X=935$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=180$ represent the population standard deviation

n=200 represent the sample size

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

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

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=3278.222$

The sample deviation calculated $s=97.054$

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 $z_{\alpha/2}=1.96$