Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval estimate for the mean number of calories for servings of breakfast cereals is (195.3, 215.7).
<h3>What is a t-distribution confidence interval?</h3>
The confidence interval is:

In which:
is the sample mean.- t is the critical value.
- n is the sample size.
- s is the standard deviation for the sample.
From the sample and the significance level of 0.05, we have that the parameters are given by:

Hence:


The 95% confidence interval estimate for the mean number of calories for servings of breakfast cereals is (195.3, 215.7).
More can be learned about the t-distribution at brainly.com/question/16162795
Answer:
x = 2.2
Step-by-step explanation:
-0.45x + 0.33 = -0.66
Subtract .33 from each side
-0.45x + 0.33-.33 = -0.66-.33
-.45x = -.99
Divide each side by -.45
-.45x./-.45 = -.99/-.45
x = 11/5
x = 2.2
Answer:
k = ⅕
Step-by-step explanation:
The slope-intercept equation for a straight line is
y = mx + b, where
m = the slope and
b = the y-intercept
Data:
(3,4) = a point on the line
(3k,0) = x-intercept
(0,-5k) = y-intercept
Calculations:
1. Slope
m = (y₂ - y₁)/(x₂ - x₁) = (-5k - 0)/(0 - 3k) = -5/(-3) = ⁵/₃
This makes the equation
y = ⁵/₃x - 5k
2. k
Insert the value of the known point: (3,4)
4 = (⁵/₃)(3) - 5k
4 = 5 - 5k
-1 = -5k
k = ⅕
The figure below shows your graph passing through (3,4) with intercepts 3k and -5k on the x- and y-axes respectively
.
Answer:
The p value for this case would be given by:
For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 0.1 and then Company B can reject the shipment
Step-by-step explanation:
Information provided
n=400 represent the random sample taken
X=59 represent number of defectives from the company B
estimated proportion of defectives from the company B
is the value to verify
represent the significance level
z would represent the statistic
represent the p value
Hypothesis to test
We want to verify if the true proportion of defectives is higher than 0.1 then the system of hypothesis are.:
Null hypothesis:
Alternative hypothesis:
The statistic would be given by:
(1)
Replacing the info given we got:
The p value for this case would be given by:
For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 0.1 and then Company B can reject the shipment