Answer:
a) 74.0353 < u < 74.0367
b) 74.036 < u
Step-by-step explanation:
Given:
- Sample size n = 15
- Standard deviation s.d = 0.0001 in
- Sample mean x_bar = 74.036
Find:
A) Calculate the 99% two-sided confidence interval on the true mean piston diameter.
Solution:
- For 99% CI, We have a = 0.01 , and Z_a/2 = 2.58
------> (x_bar - Z_a/2*s.d / sqrt(n)) < u < (x_bar + Z_a/2*s.d / sqrt(n))
------> (74.036 - 2.58*0.0001 / sqrt(15)) < u < (74.036 + 2.58*0.0001 / sqrt(15))
------> 74.0353 < u < 74.0367
Find:
B) Calculate the 95% one-sided lower confidence interval on the true mean piston diameter.
Solution:
- For 95% lower CI, We have a = 0.05 , and Z_a = 1.645
------> (x_bar - Z_a/2*s.d / sqrt(n)) < u
------> (74.036 - 1.645*0.0001 / sqrt(15)) < u
------> 74.036 < u
3 goes into 5 one time
Hope this helped :)
Answer:32p+24
Step-by-step explanation:
-4(-9p + -3) +2(-2p + 6)
36p+12-4p+12
32p+24
Hope it helped you
4^-5=1/1024 or 2/2048 or 3/3072
Answer:
The correct option is;
c. Because the p-value of 0.1609 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of juice in all the bottles filled that day does not differ from the target value of 275 milliliters.
Step-by-step explanation:
Here we have the values
μ = 275 mL
275.4
276.8
273.9
275
275.8
275.9
276.1
Sum = 1928.9
Mean (Average), = 275.5571429
Standard deviation, s = 0.921696159
We put the null hypothesis as H₀: μ₁ = μ₂
Therefore, the alternative becomes Hₐ: μ₁ ≠ μ₂
The t-test formula is as follows;

Plugging in the values, we have,
Test statistic = 1.599292
at 7 - 1 degrees of freedom and α = 0.05 = ±2.446912
Our p-value from the the test statistic = 0.1608723≈ 0.1609
Therefore since the p-value = 0.1609 > α = 0.05, we fail to reject our null hypothesis, hence the evidence suggests that the mean does not differ from 275 mL.