Question

Problem(7) Let z Student's t-distribution. The density function of T is Normal(0, 1) and Y ~ xã, then the new r.v. T = Jun has the r[(y + 1)/2) -(4+1)/2 fr(t) = + (7/2) (a) (3 points) Describe the similarity/difference between T and Z. (b) (6 points) Let to be a particular value of t. Use t-distribution table to find to values such that the following statements are true. (Given that df=10.) (i) P(-to <t<to) = 0.90. (ii) Pt S-to or t > to) = 0.05. (iii) P(t <to) = 0.025. (c) (4 points) The following random sample was selected from a normal distribution: 11 11 4,6,3,5,9, 3, 4, 4,7,8,5. Hint: &: = 58; x = 346 (i) Construct a 90% confidence interval for the population H. (ii) Construct a 95% confidence interval for the population p.

(b) and (c) are what i need help with

Answer 1

(b) The t-table I will use has computes the probabilities .

(i)

using symmetry of t curve

As it is given that

P(-to <t<to) = 0.90 →1-2P(t <-to) = 0.90 P(t <-to) = 0.05

From the table we get

(ii)  using symmetry of t curve we have

P(t <-to or t > to) = 2P(t <-to)

As

$P(t\leq -t_0 \text{ or } t>t_0) =0.05 \\ \Rightarrow P(t\leq -t_0) =0.025 \\ \Rightarrow t_0= 2.2281$

(iii) Here

c) The sample mean and sample standard deviation is $\bar{x}={1\over 11}\sum _{i=1}^{n}x_i \approx 5.273$ $s=\sqrt{{\sum_{i=1}^{n}(x_i-\bar{x})^2 \over n-1}}\approx 2.005$

(i) The $(1-\alpha)100\%$ confidence interval is given by

$\left ( \bar{x}-{s\over \sqrt{11}}t_0~,~ \bar{x}+{s\over \sqrt{11}}t_0 \right )= \left ( 5.273-{2.005\over \sqrt{11}}(1.812)~,~ 5.273+{2.005\over \sqrt{11}}(1.812) \right ) \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(4.177~,~ 6.369 )$

Use b) (i) part

(ii) Here the interval is

$\left ( \bar{x}-{s\over \sqrt{11}}t_0~,~ \bar{x}+{s\over \sqrt{11}}t_0 \right )= \left ( 5.273-{2.005\over \sqrt{11}}(2.2281)~,~ 5.273+{2.005\over \sqrt{11}}(2.2281) \right ) \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(3.926 ~,~6.620 )$