Express 1010two to base ten

A manufacturer of metal fasteners expects to ship an average of 1197.00 boxes of fasteners per day. A random sample of 24 days p

lana66690 [7]

Answer:

The p-value obtained is less than the significance level at which the test was performed, hence, we reject the null hypothesis, accept the alternative hypothesis & conclude that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Step-by-step explanation:

For hypothesis testing, we first clearly state our null and alternative hypothesis.

The null hypothesis is that there is no evidence showing that the average number of boxes shipped per day is different from 1197.00. That is, there is no significant difference between the number of boxes shipped per day and 1197.00

And the alternative hypothesis is that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Mathematically, the null hypothesis is

H₀: μ₀ = 1197.00

The alternative hypothesis is

Hₐ: μ₀ ≠ 1197.00

To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ₀)/σₓ

x = sample mean = 1193.69 boxes per day

μ₀ = standard that we're comparing the sample mean with = 1197.00

σₓ = standard error of the sample mean

= (σ/√n)

where n = Sample size = 24 days

σ = standard deviation of the sample = 3.3166 boxes per day

σₓ = (3.3166/√24) = 0.677

t = (1193.69 - 1197) ÷ 0.677

t = -4.89

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 24 - 1 = 23

Significance level = 0.05

p-value (for t = -4.89, at 0.05 significance level, with a two tailed condition) = 0.000061

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 5% = 0.05

p-value = 0.000061

0.000061 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & conclude that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Hope this Helps!!!

6 0

2 years ago

Show that the sets of upper-triangular and lower-triangular matrices are subspaces of F n,n . Call them U and L respectively. Wh

soldi70 [24.7K]

Answer:

dim L = dim U = $\frac{n(n+1)}{2}$

Step-by-step explanation:

We can do it only for the lower-triangular matrices, the case of the upper-triangular matrices is similar. We might caracterice nxn the lower-triangular matrices, as the nxn matrices $A=(a_{ij})$ such that the entry $a_{ij}=0$ if i<j.

Now let $A=(a_{ij})$ and $B=(b_{ij})$ be two lower triangular matrices, now if

$C=A+\lambda B$ for some $\lambda \in \mathbb{F}$

then the entry $c_{ij}$ of C is equal to

$c_{ij}=a_{ij}+\lambda b_{ij}$

Now, if i<j, it must hold that $a_{ij}=0 \quad \text{and} \quad b_{ij}=0$ . Therefore, if this is the case we must have that $c_{ij}=0$ and so we get that C is also a lower triangular matrix. This showa that L is closed under sum and scalar multiplcation, hence it is a linear subspace.

To find the dimension, note that all the entries of a lower-triangular matrix over the diagonal must be equal to zero. However, each entry of the matrix under the diagonal and in the diagonal might be any element of $\mathbb{F}$ , any entry that can be choosen add up to the dimension of L, we n such elemnts for the first column, (n-1) for the second column, (n-2) for the third column etc.... Therefore,