This is due today PLSS help!

What number is in between 3.75 and 4.00?

natulia [17]

3.76, 3.77, 3.78, 3.79, 3.80, 3.81, 3.82, 3.82, 3.83,

5 0

3 years ago

In constructing a confidence interval estimate for a population mean, when we replace S with the sample standard deviation (s),

Akimi4 [234]

Answer:

D

Step-by-step explanation:

When the population standard deviation is unknown while constructing the confidence interval then the estimate of standard deviation calculated from sample is used and due to this the new type of variability arise and then for conducting a confidence interval for mean the t-distribution is used. For calculating confidence interval for mean there are two sampling distributions z distribution and t distribution. When sample size is small and population standard deviation is unknown then t-distribution is used.

7 0

3 years ago

8 (3g+2)-3g=3 (5g-4)-2

serious [3.7K]

8 (3g+2)-3g=3 (5g-4)-2
Multiply the number outside the exponent with the numbers inside the parenthisis.
24g+16-3g=15g-12-2
Subtract 3g from 24g
21g+16=15g-12-2
Add 15g to 21g
36g+16=-12-2
Subtract 2 from -12
36g+16=-14
Subtract 16 from -14
36g=30
Final Answer: g=0.83 (The three goes on infinitely.)

7 0

3 years ago

The accounting department analyzes the variance of the weekly unit costs reported by two production departments. A sample of 16

dusya [7]

Answer:

$F=\frac{s^2_2}{s^2_1}=\frac{5.5}{2.5}=2.2$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have $n_2 -1 =16-1=15$ and for the denominator we have $n_1 -1 =16-1=15$ and the F statistic have 15 degrees of freedom for the numerator and 15 for the denominator. And the P value is given by:

$p_v =2*P(F_{15,15}>2.2)=0.138$

For this case the p value is highert than the significance level so we haev enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviations are not significantly different

Step-by-step explanation:

Information given

$n_1 = 16$ represent the sampe size 1

$n_2 =16$ represent the sample 2

$s^2_1 = 2.5$ represent the sample deviation for 1

$s^2_2 = 5.5$ represent the sample variance for 2

$\alpha=0.10$ represent the significance level provided

The statistic is given by:

$F=\frac{s^2_2}{s^2_1}$

Hypothesis to test

We want to test if the variations in terms of the variance are equal, so the system of hypothesis are:

H0: $\sigma^2_1 = \sigma^2_2$

H1: $\sigma^2_1 \neq \sigma^2_2$

The statistic is given by:

$F=\frac{s^2_2}{s^2_1}=\frac{5.5}{2.5}=2.2$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have $n_2 -1 =16-1=15$ and for the denominator we have $n_1 -1 =16-1=15$ and the F statistic have 15 degrees of freedom for the numerator and 15 for the denominator. And the P value is given by:

$p_v =2*P(F_{15,15}>2.2)=0.138$

For this case the p value is highert than the significance level so we haev enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviations are not significantly different

5 0

3 years ago

Evaluate -6-(-6) +7+(-4) + (-1).

frutty [35]

solution :--

= -6 + 6 + 7 - 4 - 1

= 6 + 7 - 6 - 4 - 1

= 13 - 11

= 2

3 0

3 years ago