A weekly salary is $810. What is the annual and semimonthly salary?
1 answer:
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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.
Question not correct, so i have attached the correct question.
Answer:
SE = 0.59
Step-by-step explanation:
The mean of the students height is;
x' = (53 + 52.5 + 54 + 51 + 50.5 + 49.5 + 48 + 53 + 52 + 50)/10
x' = 51.35
Now, deviation from the mean for each height;
53 - 51.35 = 1.65
52.5 - 51.35 = 1.15
54 - 51.35 = 2.65
51 - 51.35 = -0.35
50.5 - 51.35 = -0.85
49.5 - 51.35 = -1.85
48 - 51.35 = -3.35
53 - 51.35 = 1.65
52 - 51.35 = 0.65
50 - 51.35 = -1.35
Now, square of the deviations above;
1.65² = 2.7225
1.15² = 1.3225
2.65² = 7.0225
-0.35² = 0.1225
-0.85² = 0.7225
-1.85² = 3.4225
-3.35² = 11.2225
1.65² = 2.7225
0.65² = 0.4225
-1.35² = 1.8225
Sum of the squared deviations;
2.7225 + 1.3225 + 7.0225 + 0.1225 + 0.7225 + 3.4225 + 11.2225 + 2.7225 + 0.4225 + 1.8225 = 31.525
Let's divide the sum by the DF of n - 1 i.e 10 - 1 = 9.
Thus;
31.525/9 = 3.50278
Taking the square root of that gives us the standard deviation.
Thus;
s = √3.50278
s = 1.8716
Formula for standard error is;
SE = s/√n
SE = 1.8716/√10
SE = 0.59
