Can someone please help me

The proportion of defective computers built by Byte Computer Corporation is 0.15. In an attempt to lower the defective rate, the

FinnZ [79.3K]

Answer:

a) Null hypothesis : H₀: the proportion of defective item of computer has been lowered. That is P < 0.15

Alternative hypothesis: H₁: The proportion of defective item of computer

has been higher. That is P> 0.15 (Right tailed test)

b) Test statistic $Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }$

c) Calculate the value of the test statistic = 0.991

d) The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57

e) Null hypothesis accepted at 0.01 level of significance

f) we accepted null hypothesis.

Hence the proportion of defective item of computer has been lowered.

Step-by-step explanation:

Step(i):-

Given the sample size 'n' = 42

Given random sample of 42 computers were tested revealing a total of 4 defective computers.

The defective computers 'x' = 4

The sample proportion of defective computers

$p = \frac{x}{n} = \frac{4}{42} = 0.095$

Given The Population proportion 'P' = 0.15

The level of significance ∝=0.01

Step(ii):-

a) Null hypothesis : H₀: the proportion of defective item of computer has been lowered. That is P < 0.15

Alternative hypothesis: H₁: The proportion of defective item of computer

has been higher. That is P> 0.15 (Right tailed test)

b)

Test statistic $Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }$

c)

$Z = \frac{0.095-0.15}{\sqrt{\frac{0.15(0.85)}{42} } }$

$z = \frac{-0.055}{\sqrt{0.00303} } = - 0.9991$

Calculate the value of the test statistic Z = - 0.9991

|Z| = |- 0.9991| = 0.991

Step(iii):-

d)

The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57

e) Calculate the value of the test statistic Z = 0.991 < 2.57 at 0.01 level of significance.

Conclusion:-

Hence the null hypothesis is accepted at 0.01 level of significance.

f)

The proportion of defective item of computer has been lowered.

6 0

3 years ago

miles wants some new games for his phone. each game costs 2.99 how much would it cost if he bought 7 new games

bekas [8.4K]

Answer: $20.93

Step-by-step explanation:

If you take a game that costs $6.99, and then multiplied that by 7, you would get $20.93.

Let me know if this helps!!!

:D

6 0

3 years ago

Read 2 more answers

This graph shows the altitude of an airplane over time. Which story matches the graph?

Oksi-84 [34.3K]

Answer:

A.) The aircraft rose quickly into the air at takeoff, and then continued at a constant altitude.:)

6 0

3 years ago

Read 2 more answers

What are expressions like 4(3+2) and 4(3)+4(2)

romanna [79]

Your using the distributive property in your example, hope that helps.

5 0

3 years ago

(CO6) From a random sample of 68 businesses, it is found that the mean time that employees spend on personal issues each week is

satela [25.4K]

Answer:

(1) (4.82, 4.98)

(2) Large sample size

(3) Yes, the temperature is within the confidence interval of (37.40, 37.60)

(4) (15.083, 15.117)

Step-by-step explanation:

Confidence Interval (CI) = mean + or - (t×sd)/√n

(1) mean = 4.9, sd = 0.35, n = 68, degree of freedom = n-1 = 68 - 1 = 67

t-value corresponding to 67 degrees of freedom and 95% confidence level is 1.9958

CI = 4.9 + (1.9958×0.35)/√68 = 4.98

CI = 4.9 - (1.9958×0.35)/√68 = 4.82

CI is (4.82, 4.98)

(2) Error margin = (t-value × standard deviation)/√sample size

From the formula above, error margin varies inversely as the square root of the sample size. Since the relationship between the error margin and sample size is inverse, increase in one (sample size) will conversely lead to a decrease in the other (error margin)

(3) mean = 37.5, sd = 0.6, n= 100, degree of freedom = n-1 = 100-1 = 99

t-value corresponding to 99 degrees of freedom and 90% confidence level is 1.6602

CI = 37.5 + (1.6602×0.6)/√100 = 37.5 + 0.10 = 37.60

CI = 37.5 - (1.6602×0.6)/√100 = 37.5 - 0.10 = 37.40

37.53 is within the confidence interval (37.40, 37.60)

(4) mean = 15.10, sd =0.08, n = 104, degree of freedom = n-1 = 104-1 = 103

t-value corresponding to 103 degrees of freedom and 97% confidence interval is 2.2006

CI = 15.10 + (2.2006×0.08)/√104 = 15.10 + 0.017 = 15.117

CI = 15.10 - (2.2006×0.08)/√104 = 15.10 - 0.017 = 15.083

CI is (15.083, 15.117)

4 0

3 years ago