हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी वाणिज्य कक्षा १२

The mean breaking strength of cables supplied by a manufacturer is 1,800 with a standard deviation of 100. By a new technique in the manufacturing process it is claimed that the break - Business Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

The mean breaking strength of cables supplied by a manufacturer is 1,800 with a standard deviation of 100. By a new technique in the manufacturing process it is claimed that the breaking strength of the cables has increased. In order to test this claim a sample of 50 cables is tested. It is found that the mean breaking strength is 1,850. Can you support the claim at 0.01 level of significance?

योग

उत्तर

Sample size n = 50

Sample mean `bar(x)` = 1800

Sample SD S = 100

Population mean µ = 1850

Null hypothesis H0: µ = 1850

Level of significance µ = 0.01

Test statistic:

z = `(bar(x) - mu)/(sigma/sqrt("n")) "N" ∼ (0.1)`

z = `(1800 - 1840)/((100/sqrt(50))` 

= `(-50)/sqrt(100/(7.071)`

= `(-50 xx 7.0711)/100`

= `7.0711/2`

= – 3.5355

∴ z = – 3.536

Calculated value |z| = 3.536

Critical value at 1% level of significance is `"Z"_(("a")/2)` = 2.58

Inference:

Since the calculated value is greater than table

i.e. `"Z"_(("a")/2)` at 1% level of significance, the null hypothesis is rejected

Therefore we conclude that we is rejected.

Therefore we conclude that we can not support we conclude that we can support the claim of 0.01 of significance.

shaalaa.com
Hypothesis Testing
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Sampling techniques and Statistical Inference - Exercise 8.2 [पृष्ठ १९५]

APPEARS IN

सामाचीर कलवी Business Mathematics and Statistics [English] Class 12 TN Board
अध्याय 8 Sampling techniques and Statistical Inference
Exercise 8.2 | Q 17 | पृष्ठ १९५
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×