What is the range of this function?

What are the critical values for a two-tailed test with a 0.01 level of significance when n is large and the population standard

Sunny_sXe [5.5K]

For a two-tailed test we need to halve 0.01, giving 0.005. Then we need to sutract 0.005 from 0.5, giving 0.5 - 0.005 = 0.495. Looking up the p value of 0.495 in an inverse normal probability table we obtain the value z = 2.5758. Therefore the critical values are -2.5758 and 2.5758.

4 0

3 years ago

A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar

Arisa [49]

Answer:

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

$n=17$ represent the sample selected

$\alpha=0.1$ represent the significance

$s^2 =4.7^2 =22.09$ represent the sample variance

$\sigma^2_0 =3.5^2 =12.15$ represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 12.15$

Alternative hypothesis: $\sigma^2 \neq 12.15$

The statistic for this case is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

The degrees of freedom are:

$df =n-1= 17-1=16$

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

7 0

3 years ago

Work out the height of a parallelogram with a base of 10m and an area of 60m2

noname [10]

Area formula of a parallelogram is
A = base x height

Given are the
Area = 60 m² and base = 10 m

Substitute the given:

60 = 10h (divide by 10 to find value of h)
$\frac{60}{10} = \frac{10h}{10}$ (10 and 10 cancels out)

h = 6
The height of a parallelogram with a base of 10m and an area of 60m² is 6m

3 0

3 years ago

What is 15 degrees less than 78.6 degrees?

Jobisdone [24]

Answer:

15 degrees less --> substract 15 degrees.

78.6 - 15 = 63.6 degrees

7 0

3 years ago

Read 2 more answers

I need the answers to 15,16,17,18

Mila [183]

Well for 17 it’s 125%

8 0

2 years ago

What is the range of this function?​

What is the range of this function?