In the hundreds spot is a 2 and 2 is under 5 so we round down which would be 4 398 200
Answer:
Step-by-step explanation:
Outline are values which are entirely different from those remaining values in a data set. These extreme values can skew an approximately normal distribution by skewing the distribution in the direction of the outliers and this makes it difficult for the data set to be analyzed.
Its effect is such that the mean becomes extremely sensitive to extreme outliers making it possible that the mean is this not a representative of the population and this theoretically affects the standard deviation.
Answer:
We reject H₀. We support that the new average credit card debt is bigger than the previous average
Step-by-step explanation:
Five years earlier
μ = 8189
σ = 690
Sample size n = 32
Recent year debt
x = 8776
Sample size n = 32
a) Hypothesis Test:
Null Hypothesis H₀ x = μ = 8189
Alternative Hypothesis Hₐ x > μ
b) z(c) Alternative Hypothesis establishes that the test is a one tail-test to the right.
z(c) for significance level α = 0.05 is from z-table z(c) = 1,64
c) z(s) = ( x - μ ) / σ /√n
z(s) = ( 8776 - 8189 ) / 690 /√32
z(s) = 587 *5,66/ 690
z(s) = 4,81
d) Comparing z(c) and z(s)
z(s) > z(c) Then z(c) is in the rejection region and we reject H₀
e) We have evidence that at 95 % of confidence the new value for the debt in credit card is now bigger than the average
6x15= 90
9x20= 180
180+90= 270
270 - 25 = 245 dane has 245 stamps left
Answer:
The p-value here is 0.0061, which is very small and we have evidence that the girls' mean is higher than the boys' mean.
Step-by-step explanation:
We suppose that the two samples are independent and normally distributed with equal variances. Let
be the mean number of ring tones for girls, and
the mean number of ring tones for boys.
We want to test
vs
(upper-tail alternative).
The test statistic is
T =
where
.
For this case,
,
,
,
,
.
and the observed value is
t =
.
We can compute the p-value as P(T > 2.6309) where T has a t distribution with 20 + 20 - 2 = 38 degrees of freedom, so, the p-value is 0.0061. Because the p-value is very small, we can reject the null hypothesis for instance, at the significance level of 0.05.