Ask question

Ask question

All categories

Ilia_Sergeevich [38]

3 years ago

15

An engineering consulting firm wantedto evaluate a rivet process by measuring the formed diameter. The following data represent

the diameters (in hundredths of an inch) for a random sample of 24 rivet heads:
6.81 - 6.79 - 6.69 - 6.59 - 6.65 - 6.60 - 6.74 - 6.70 - 6.76
6.84 - 6.81 - 6.71 - 6.66 - 6.76 - 6.76 - 6.77 - 6.72 - 6.68
7.71 - 6.79 - 6.72 - 6.72 - 6.72 - 6.79 - 6.83
a) Set up a 95% confidence interval estimate of the average diameter of rivet heads (in hundredths of an inch).
b) Set up a 95% confidence interval estimate of the standard deviation of the diameter of rivet heads (in hundredths of an inch)

1 answer:

AleksAgata [21]3 years ago

3 0

Answer:

Step-by-step explanation:

6.81 - 6.79 - 6.69 - 6.59 - 6.65 - 6.60 - 6.74 - 6.70 - 6.76

6.84 - 6.81 - 6.71 - 6.66 - 6.76 - 6.76 - 6.77 - 6.72 - 6.68

7.71 - 6.79 - 6.72 - 6.72 - 6.72 - 6.79 - 6.83

$\bar x =6.77$

S.D = 0.21

$I=6.77\pmt\times\frac{s}{\sqrt{n} }$

df = 24

α = 0.05

t = 2.064

$I=6.77\pm2.064\times\frac{0.21}{\sqrt{25} } \\\\=6.77\pm0.087\\\\=[6.683,6.857]$

b)

$\sqrt{\frac{(1-n)s^2}{X^2_{\alpha /2} } < \mu$

$\sqrt{\frac{24 \times 0.21^2}{39.364} } < \mu$

You might be interested in

Can anybody help me please

MatroZZZ [7]

Answer:

○ 4⁄6, -4⁄6

Step-by-step explanation:

Take the square root of both the denominator and numerator, and you will get the above two answers.

I am joyous to assist you anytime.

4 0

3 years ago

15 points please help

Sonja [21]

Answer: sorry dont know this

Step-by-step explanation:

5 0

3 years ago

If d , e, and f are midpoints of the sides of ABC, find the perimeter of ABC.

ehidna [41]

Answer:

Perimeter of the ΔDEF = 10.6 cm

Step-by-step explanation:

The given question is incomplete; here is the complete question with attachment enclosed with the answer.

D, E, and F are the midpoints of the sides AB, BC, and CA respectively. If AB = 8 cm, BC = 7.2 cm and AC = 6 cm, then find the perimeter of ΔDEF.

By the midpoint theorem of the triangle,

Since D, E, F are the midpoints of the sides AB, BC and CA respectively.

Therefore, DF ║ BC and $FD=\frac{1}{2}\times(BC)$

FD = $\frac{7.2}{2}$

= 3.6

Similarly, $FE=\frac{1}{2}\times(AB)$

$FE=\frac{8}{2}$

FE = 4 cm

And $DE=\frac{AC}{2}$

DE = $\frac{6}{2}$

= 3 cm

Now perimeter of ΔDEF = DE + EF + FD

= 3 + 4+ 3.6

= 10.6 cm

Perimeter of the ΔDEF is 10.6 cm.

7 0

3 years ago

On the Sat the lowest value anyone can score on the math section is 200 and the highest is 800. Jason is a senior retaking the S

schepotkina [342]

Answer:

10 people

Step-by-step explanation:

Parking needs to be paid for but if all friends are going together, they need to only pay once for parking.

The amount spent on just tickets is therefore:

= 243.25 - 15.75

= $227.50.

If $227.50 was for tickets alone, the number of tickets they could buy is:

= Total amount left/ Price of tickets

= 227.50 / 22.75

= 10 people

Step-by-step explanation:

6 0

3 years ago

A dataset lists full IQ scores for a random sample of subjects with low lead levels in their blood (sample 1) and another random

Alenkasestr [34]

Answer:

a. Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 >\mu_2$

b. $t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282$

c. $p_v =P(t_{97}>2.287) =0.0122$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

Step-by-step explanation:

a. State and label the null and alternative hypotheses.

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 >\mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 \leq 0$

Alternative hypothesis: $\mu_1 -\mu_2>0$

Our notation on this case :

$n_1 =78$ represent the sample size for group 1

$n_2 =21$ represent the sample size for group 2

$\bar X_1 =92.88$ represent the sample mean for the group 1

$\bar X_2 =86.90$ represent the sample mean for the group 2

$s_1=15.34$ represent the sample standard deviation for group 1

$s_2=8.99$ represent the sample standard deviation for group 2

b. State the value of the test statistic.

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom. If we replace the values given we have:

$t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282$

Now we can calculate the degrees of freedom given by:

$df=78+21-2=97$

c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Method used: P value

And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:

$p_v =P(t_{97}>2.287) =0.0122$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

6 0

3 years ago