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13. Based on a random sample of size 20 from a normal distribution with variance o?,...
13. Based on a random sample of size 20 from a normal distribution with variance o?,...
13. Based on a random sample of size 20 from a normal distribution with variance o?, the width of the 95% confidence interval for ois 150. Let xã, be the the critical value of a chi-squared random variable with v degrees of free- dom. The following table lists values of x. for specific combinations of a and v: v = 19 v = 20 a = 0.975 8,907 9.591 a = 0.95 10.117 10.851 a = 0.05 30.144 3 1.410...
SAT test scores are reported to follow a normal distribution with variance 125. A re- searcher...
SAT test scores are reported to follow a normal distribution with variance 125. A re- searcher questions if the assumed variance ought to be lower, and randomly samples 20 test scores. The sample variance of the 20 scores is 104. = 0.95a = 0.05 = 0.025 a0.975 30.144 v19 8.907 10.117 32.852 V = 20 9.591 10.851 31.410 34.170 Test whether the researcher is justified at the 5% significance level
10. Based on a random sample of size 7 from a normal distribution with mean y,...
10. Based on a random sample of size 7 from a normal distribution with mean y, a confidence interval is constructed for y. The sample standard deviation is calculated as 5.4763. Let to be the critical value of a t random variable with v degrees of freedom. The following table lists values of ta for specific combinations of a and v: v=6 v = 7 a=0.1 a = 0.05 a= 0.025 1.440 1.943 2 .447 1.4151.8952.365 If we want to...
10. Based on a random sample of size 7 from a normal distribution with mean u,...
10. Based on a random sample of size 7 from a normal distribution with mean u, a confidence interval is constructed for y. The sample standard deviation is calculated as 5.4763. Let tay be the critical value of a t random variable with v degrees of freedom. The following table lists values of ta, for specific combinations of a and v: a=0.1 1.440 1.415 a= 0.05 1.943 v=6 v=7 a= 0.025 2 .447 2.365 1.895 If we want to be...
you are given For a random sample of size 13 from a normal distribution with mean...
you are given For a random sample of size 13 from a normal distribution with mean the following regarding the observations: 3 (xi – ī)2 = 77.8 The width of the 100k% confidence interval for u is 2.7005. Let to be the critical value of a t random variable with v degrees of freedom. The following table lists values of ta, for specific combinations of a and v: v = 12 v = 13 a=0.1 1.356 1.350 a= 0.07 1...
11. For a random sample of size 13 from a normal distribution with mean u, you...
11. For a random sample of size 13 from a normal distribution with mean u, you are given the following regarding the observations: (ti – 1)2 = 77.8 The width of the 100% confidence interval for u is 2.7005. Let tay be the critical value of a t random variable with v degrees of freedom. The following table lists values of tay for specific combinations of a and v: v = 12 v=13 a=0.1 1.356 1.350 a= 0.07 1.580 1.572...
Consider a random sample from a normal population with mean u = 3 and variance o2...
Consider a random sample from a normal population with mean u = 3 and variance o2 = 22, with sample size n = 20. Suppose the sample variance is 82 = 2.72. Let p be the probability that s2 exceeds the sample variance 52. Which of the following is true? 0.01 < p < 0.025 0.025 < p < 0.05 0.05<P 0.005< p < 0.01 Op < 0.005
Consider a random sample from a normal population with mean u = 3 and variance o2...
Consider a random sample from a normal population with mean u = 3 and variance o2 = 22, with sample size n = 20. Suppose the sample variance is 82 = 2.72. Let p be the probability that s2 exceeds the sample variance 52. Which of the following is true? 0.01 < p < 0.025 0.025 < p < 0.05 0.05<P 0.005< p < 0.01 Op < 0.005
Consider a random sample from a normal population with mean u = 3 and variance o2...
Consider a random sample from a normal population with mean u = 3 and variance o2 = 22, with sample size n = 20. Suppose the sample variance is 82 = 2.72. Let p be the probability that s2 exceeds the sample variance 52. Which of the following is true? 0.01 < p < 0.025 0.025 < p < 0.05 0.05<P 0.005< p < 0.01 Op < 0.005
Let X1, X2, ..., Xn be a random sample from the N(u, 02) distribution. Derive a...
Let X1, X2, ..., Xn be a random sample from the N(u, 02) distribution. Derive a 100(1-a)% confidence interval for o2 based on the sample variance S2. Leave your answer in terms of chi-squared critical values. (Hint: We will show in class that, for this normal sample, (n − 1)S2/02 ~ x?(n − 1).)
13. Based on a random sample of size 20 from a normal distribution with variance o?, the width of the 95% confidence interval for ois 150. Let xã, be the the critical value of a chi-squared random variable with v degrees of free- dom. The following table lists values of x. for specific combinations of a and v: v = 19 v = 20 a = 0.975 8,907 9.591 a = 0.95 10.117 10.851 a = 0.05 30.144 3 1.410...
SAT test scores are reported to follow a normal distribution with variance 125. A re- searcher questions if the assumed variance ought to be lower, and randomly samples 20 test scores. The sample variance of the 20 scores is 104. = 0.95a = 0.05 = 0.025 a0.975 30.144 v19 8.907 10.117 32.852 V = 20 9.591 10.851 31.410 34.170 Test whether the researcher is justified at the 5% significance level
10. Based on a random sample of size 7 from a normal distribution with mean y, a confidence interval is constructed for y. The sample standard deviation is calculated as 5.4763. Let to be the critical value of a t random variable with v degrees of freedom. The following table lists values of ta for specific combinations of a and v: v=6 v = 7 a=0.1 a = 0.05 a= 0.025 1.440 1.943 2 .447 1.4151.8952.365 If we want to...
10. Based on a random sample of size 7 from a normal distribution with mean u, a confidence interval is constructed for y. The sample standard deviation is calculated as 5.4763. Let tay be the critical value of a t random variable with v degrees of freedom. The following table lists values of ta, for specific combinations of a and v: a=0.1 1.440 1.415 a= 0.05 1.943 v=6 v=7 a= 0.025 2 .447 2.365 1.895 If we want to be...
you are given For a random sample of size 13 from a normal distribution with mean the following regarding the observations: 3 (xi – ī)2 = 77.8 The width of the 100k% confidence interval for u is 2.7005. Let to be the critical value of a t random variable with v degrees of freedom. The following table lists values of ta, for specific combinations of a and v: v = 12 v = 13 a=0.1 1.356 1.350 a= 0.07 1...
11. For a random sample of size 13 from a normal distribution with mean u, you are given the following regarding the observations: (ti – 1)2 = 77.8 The width of the 100% confidence interval for u is 2.7005. Let tay be the critical value of a t random variable with v degrees of freedom. The following table lists values of tay for specific combinations of a and v: v = 12 v=13 a=0.1 1.356 1.350 a= 0.07 1.580 1.572...
Consider a random sample from a normal population with mean u = 3 and variance o2 = 22, with sample size n = 20. Suppose the sample variance is 82 = 2.72. Let p be the probability that s2 exceeds the sample variance 52. Which of the following is true? 0.01 < p < 0.025 0.025 < p < 0.05 0.05<P 0.005< p < 0.01 Op < 0.005
Consider a random sample from a normal population with mean u = 3 and variance o2 = 22, with sample size n = 20. Suppose the sample variance is 82 = 2.72. Let p be the probability that s2 exceeds the sample variance 52. Which of the following is true? 0.01 < p < 0.025 0.025 < p < 0.05 0.05<P 0.005< p < 0.01 Op < 0.005
Consider a random sample from a normal population with mean u = 3 and variance o2 = 22, with sample size n = 20. Suppose the sample variance is 82 = 2.72. Let p be the probability that s2 exceeds the sample variance 52. Which of the following is true? 0.01 < p < 0.025 0.025 < p < 0.05 0.05<P 0.005< p < 0.01 Op < 0.005
Let X1, X2, ..., Xn be a random sample from the N(u, 02) distribution. Derive a 100(1-a)% confidence interval for o2 based on the sample variance S2. Leave your answer in terms of chi-squared critical values. (Hint: We will show in class that, for this normal sample, (n − 1)S2/02 ~ x?(n − 1).)
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Amplitude=3;
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n=0:399;
t=0:1/fs: n*1/fs-1/fs;
signal=3+3*cos(2*pi*1100*t)+3*cos(2*pi*2200*t)+3*cos(2*pi*3300*t);
fftSignal= fft(signal);
fftSignal=f ftshift (fftSignal);
f=fs/2*linspace(-1,1,fs);
plot(f,abs(fftsignal);
xlabel('Frequency(Hz)’)
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