Tutorial 17-Hypothesis Testing And Statistical Analysis Using Z test

A Z-test is any statistical test for which the distribution of the test statistic under the null hypothesis can be approximated by a normal distribution. Z-tests test the mean of a distribution.
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[@vyasishan3800]

Thank you Krish for putting efforts to explain this concepts.

[@arpitasharma28]

Really appreciate your efforts to upload this video... great explanation sir:)

[@rohitrohit6883]

Sir, I am very thankful to you for this video.🙂😇

[@MuhammadIbrahim-ju1tp]

great krish respect so much value

[@madhuchandra123]

Please complete the statistics playlist.

[@vishaltiwaribnt]

thanks for Uploading in Hindi.....its much more understaniding.....

[@amartamboli6718]

Null Hypothesis = 5 Alternative Hypothesis = not equal to 5 Significance level = 2%,1-1% either side... Area under curve using positive z-score table = 1-0.01=0.99=2.33(As per Z-score positive table) Decision Boundary = +- 2.33 Z-score = -2.53 Since Z-test value is less than -2.33 i/e. it is in the negative rejection region , So we reject the null Hypothesis ..

[@salmanshaikh5851]

Bhai mere ko to abhi bhi palle nai pada Kaha se tune sikha batai gake?

[@hades840]

it is one tail test look there is underline on less

[@harshshukla980]

decision boundary 2.05

[@sarveshpandey9047]

2.33 kha se aya z table me to diya he nhi h

[@ashan3453]

the explanation is too good sir

[@anuraaganath]

pmean = 5, std = 0.5, smean=4.8, n=40; null hypothesis = pmean = 5 alternate hypothesis = pmean < 5 C.I = 1-0.02 = 0.98 z = -2.52 where the decision boundry is -2.06 as per the z table. So, we failed to accept the null hypothesis and the warranty should be revised

[@alex-vq1yy]

can you explain how decision boundary = -2.06 how you calculated CI please explain

[@MH_POLITICSS]

Decision boundry= +-2.33 Z score = - 2.52 Conclusion - if z score is less than -2.33 or greater than -2.33 then we reject null hypothesis. So -2.52 >-2.33 we Reject null hypothesis..

[@mahalaxmiindapure8961]

I got decision boundary between -2.33 to +2.33, z-score= -2.529, We will reject the null hypothesis means workers believe is true.

[@sumitkoshti1171]

Yes

[@Khojbysridhar]

You are amazing

[@ashishvinod2193]

since our entire area under the curve value is = 1 -0.98 = 0.02 since it's left tail test.. out z test value =-2.52 2.52

[@rajsinghtomar8090]

no confidence interval how i solve plz reply

[@ashishvinod2193]

@@rajsinghtomar8090 if confidence interval was not given then by default assume 98% confidence intervals

[@sumitkoshti1171]

@@rajsinghtomar8090 bhai significance value is given 2% which is equal to 0.02, and by using significance value you can find CI from the formula significance value = 1-CI, 0.02 = 1-CI, CI = 1-0.02 = 0.98 which is 98%

[@zarmeenakhan899]

can you make a video on cohort and case control as well?

[@sketchytv1321]

بہت آسانی سے مشکل کنسپٹ سمجھایا ۔شکریہ

[@EjazUddin786]

Hanji bilkul

[@baishalibiswas7525]

Please complete full playlist

[@hardikanandani7377]

Ho = average warranty is 5 years. H1= average warranty is not equal to 5 years. Alpha = 0.02(Here it's less than 5 years ,so it's one tailed.) Area under curve = 1-0.02 = 0.098 Decision boundry = +-2.06 Z-score = -2.52 So Ho is rejected. So, the worker was right I think. Bulb will malfunction in less than 5 years.

[@souravsinha18]

yes it is correct but decision boundary should be -2.06 not + - 2.06

[@hardikanandani7377]

Correct.

[@sumiturkude1395]

@@hardikanandani7377 how did you find decision boundary.?

[@shashi_kamal_chakraborty]

@@sumiturkude1395 It's 1 - alpha = 1 - 0.02 = 0.98. Now, z-table look up for 0.98. Which is found to be -2.06

[@KartikSharma12364]

A chocolate company claims that their new recipe for dark chocolate bars increases the average cocoa content from the previous recipe. They conducted a study where they sampled 50 dark chocolate bars and found that the sample mean cocoa content is 85%. The population standard deviation of cocoa content is known to be 8%. Test the company's claim at a significance level of α = 0.05. Solve this problem

[@user-uh6th9vc5x]

V nice Sir

[@shaitan779gaming9]

Hi Krish, is it standard that we should always go with Z-test when sample size is above 30?

[@pr_ateek]

Please complete the statistics playlist

[@rishabhchauhan8612]

This topic is For Data Analytics or Data science ??

[@HarshitSharma-df7kx]

sir what do we do when we do not have a 2 tail test condition ??

[@manjeshtiwari7434]

Thank you for such an amazing lecture

[@KartikSharma12364]

My decision boundary is not getting correct ,can you help me

[@manjeshtiwari7434]

@@KartikSharma12364 what's issue?

[@manjeshtiwari7434]

@@KartikSharma12364 if accept area is 98 how can be reject 49 and 49, out of 100 98 is accept area so we have 1% reject area on both side u need to rewatch

[@KartikSharma12364]

@@manjeshtiwari7434 what's the correct answer of decision boundary and z test

[@KartikSharma12364]

​@@manjeshtiwari7434decision boundary=2.33

[@AmitSharma-oh5uw]

make more video correlation regression and probability in stats

[@Cricstory0108]

Sir konse video ha hypothesis check karne wali

[@rajarshiroy8794]

I have a doubt, From other definitions I think that If we have the population standard deviation we can directly move for z test since we are concerned about the sample size only when we are talking about the sample standard deviation as for n>30 sample and population SD are similar. So, I think the flowchart would be a little different. Can anybody clarify this doubt please??

[@harshitgupta6452]

Yes you are correct

[@salmanshaikh5851]

Continue Sir Continue me sahi time per aaya hu bcz mera abhi 15 ma tutorial chalra hai😂🤣

[@user-vb5hh7rw3s]

for decision boundry i use this : for 2% significane 1-1 on both sides 1/100=0.01 , 1-0.01=0.99 and in z table value of 0.99 is 2.33 means +2.33 on right and -2.33 on left pls tell me am i correct ?

[@rajatjuneja8367]

yes sir

[@alphapoint389]

Sir how formulas are derived?

[@KartikSharma12364]

A chocolate company claims that their new recipe for dark chocolate bars increases the average cocoa content from the previous recipe. They conducted a study where they sampled 50 dark chocolate bars and found that the sample mean cocoa content is 85%. The population standard deviation of cocoa content is known to be 8%. Test the company's claim at a significance level of α = 0.05. Anyone solve

[@Vishal_19]

U = 5 sd = 0.50 n = 40 x bar = 4.8 so area = 1 - 0.02 = 0.98 z_value will be 2.06 (dission boundry) 4.8 - 5 ----------- 0.05 * root 40 = - 2.52 we reject the null hypotheses the worker was right

[@swapnil2881]

Sir please stat ki playlist continue rakho

[@rajasaad3450]

wait for course completion !

[@kdasurrenderedsoul7613]

Hare Krishna 😊 Thank you so much for this vidoe 😊😊😊

[@shailendrayadav6937]

Here bulb manufacturers question is a one tail test question because , worker believe that bulb will malfunction in less than 5yrs . So Null hypothesis= H0 = 5yrs alternate hypothesis = H1

[@dishantsonawane8510]

Why do we consider sample size 30 for differentiating between Z test and T test. Why not some other number ? Why not 'x' percentage of the population ?

[@sauravjha1714]

Null Hypothesis = Mean = 5 Alternative Hypothesis = Mean not equal to 5 Significance level = 2%,1-1% either side... Area under curve using positive z-score table = 1-0.01=0.99=2.33(As per Z-score positive table) Decision Boundary = +- 2.33 Z-score = -2.52 Since Z-score falls in the negative rejection area...we reject the null Hypothesis!!!

[@rokhaiyasultana9728]

i got the sme ans!!

[@souravsinha18]

No it is not a 2 tailed test so alternative hypotheis should be mean less than 5 years

[@salmanshaikh5851]

Bhai mere ko to abhi bhi palle nai pada kaha se tune sikha pata gake?

[@sauravjha1714]

@@souravsinha18 check your answer... My answer is perfectly right!!

[@souravsinha18]

@@sauravjha1714 no bro I did this I am sure this is right why are u taking 1-1% on both sides when the worker says that the bulb lifetime will decrease you should take it on decreasing side and then calculate it

[@dhananjaykharat4812]

(Z Test = -2.53)H0 has been rejected as (Z test VAlue is not in C.I range/Decision Boundry) so worker believe is correct.

[@vaishhh_likhii]

Thankyou so muchh,such a nice explaination!! best video for Hypothesis testingg......🤌

[@wewesusui9528]

Sir anova test pr video banao baki ke saare test statistic 7 days live me hue he cover pr anova test nahi hua he

[@heeralalchoudhary1587]

Fabulous sir ..can I contact personally

[@xero4479]

Sir kal paper hai aur ye pura samajh hi nahi aa raha tha, thank you🙏

[@fehmeedaayoub6559]

Paired and unpaired ka concept????

[@Liku108]

Sir what about f test

[@hardsoul00]

Sir CHI SQUARE AND ANNOVA KA VIDEO Bna dein mera paper ha iska next month 10th november ko

[@godgdfhackergod3836]

please continue video on ml

[@akashshakya8740]

pls provide lecture notes

[@user-wn6tv1wm7u]

Is This playlist enough to learn statistics for data analytics ?

[@Prashantchaudhary-kf1fb]

yes bro

[@dilipgyawali1776]

if population standard deviation is given and sample size is less than 30, then Z test???please clarify.

[@robinsunuwar6852]

you should go with t test if the sample size will be less than 30 we have to perform t test its basic rule.

[@vishujaiswal6160]

next video pl. sir

[@zero_analyst]

Hello sir thank you for explaining You have mentioned in conclusion if z test is less than 1.96 or greater than 1.96 then we reject Null Hypothesis but as results came 2.31 which is greater than 1.96 so we should accept the Null Hypothesis. I am now confused can you please clarify thanks in advance

[@itshimanshucreation8076]

Bro unhone bola hai ki we accept to reject the H0 if z test is lessthen or greater then 1.96.

[@vedchaudhari6587]

if it lies in range of 1.96 and -1.96 then accept it else for anything either greater or smaller , reject it.

[@urvashirai8441]

If std not given and sample size more thn 30,thn which test is to be used?

[@CrocksJam]

if population std mean not given then we use t-test directly

[@theentertainmentworld3310]

Do any one get the z test value?? I'm getting my z test value which is +0.037 and decision boundary values as +3.1 and -3.1 , So i have to accept the null hypothesis Do anyone get any answer??? Sir am i correct??

[@sumitkoshti1171]

Nope, Decision boundary is +-2.33 and Z test value is -2.53, As -2.53

[@muhammadansar3373]

According to my calculation HO is accepted

[@prashindu]

For a two-tailed test with a 98% confidence interval, you would use 0.01 (or 1%) for the purpose of arriving at the z-score. This is because the remaining 2% is split equally between the two tails of the distribution, leaving 1% in each tail. So, you’d look up the z-score that corresponds to an area of 0.99 in a standard normal distribution table. However, for a one-tailed test with a 98% confidence interval, you would use 0.02 (or 2%) to find the z-score. This is because the entire 2% is located in one tail of the distribution. So, you’d look up the z-score that corresponds to an area of 0.98 in a standard normal distribution table. When in doubt, it’s generally safer to use a two-tailed test. This is because a two-tailed test is more conservative and does not assume a direction of the effect.

[@danielrichard7575]

"promo sm" 🤗

[@robotsj2775]

Worst tutorial for z test This should not take more than 7-8 mins. Pls don't go into so much detail. Instead you would have solved 4-5 problems in 25 min.

[@importantforlifeyasharora9042]

Actuall fanda ya to kisi ko pata nahi hai ya wo samjhane me inko takleef hai. Maths bhi theory ki tarah padha rahe hai😂