SOMEONE PLEASE HELP I will give you brainliest it’s due in 11:59 pm TODAY!! Please
2 answers:
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Answer:
a)
y = 0.3035 + 0.0082x
b)
0.6315 mm
c)
x = 23.9634 °C
Step-by-step explanation:
a. Compute the least-squares line for predicting warping from temperature. Round the answers to four decimal places.
We need to find an equation of the form
y = b + mx
where m is the slope and b the Y-intercept.
The slope m can be computed with the formula
Replacing the values in our formula (we will round at the end of the calculations)
the Y-intercept b is computed with the formula
therefore we have
and the least-squares line rounded to 4 decimals would be
y = 0.3035 + 0.0082x
b. Predict the warping at a temperature of 40°C. Round the answer to three decimal places.
We simply replace x with 40 to get
y = 0.3035 + 0.0082*40 = 0.6315 mm
c. At what temperature will we predict the warping to be 0.5 mm? Round the answer to two decimal places
Here, we replace y with 0.5 and solve for x
0.5 = 0.3035 + 0.0082x ===> x = (0.5-0.3035)/0.0082 ===>
x = 23.9634 °C
Answer:
a: 28 < µ < 34
Step-by-step explanation:
We need the mean, var, and standard deviation for the data set. See first attached photo for calculations for these...
We get a mean of 222/7 = 31.7143
and a sample standard deviation of: 4.3079
We can now construct our confidence interval. See the second attached photo for the construction steps.
They want a 90% confidence interval. Our sample size is 7, so since n < 30, we will use a t-score. Look up the value under the 10% area in 2 tails column, and degree of freedom is 6 (degree of freedom is always 1 less than sample size for confidence intervals when n < 30)
The t-value is: 1.943
We rounded down to the nearest person in the interval because we don't want to over estimate. It said 28.55, so more than 28 but not quite 29, so if we use 29 as the lower limit, we could over estimate. It's better to use 28 and underestimate a little when considering customer flow.
